Class 11

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(a)

Question 1.
Multiply (2√-3 + 3√-2) by (4√-3 – 5√-2)
Solution:
(2√-3 + 3√-2) by (4√-3 – 5√-2)
= (2√3i + 3√2i) (4√3i – 5√2i)
= i² (2√3 + 3√2) (4√3 – 5√2)
= – 1(24 – 10√6 + 12√6 – 30)
= – 1(- 6 + 2√6) = 6 – 2√6

Question 2.
Multiply (3√-7 – 5√-2) (3√-2 + 5√-2)
Solution:
(3√-7 – 5√-2) (3√-2 + 5√-2)
= (3√7i – 5√-2i) (3√2i + 5√2i)
= i² (3√7- 5√2) (3√2 + 5√2)
= (- 1)8√2(3√7 – 5√2 )
= – 24√14 + 80

Question 3.
Multiply (√-1 +√-1) (a – b√-1)
Solution:
(√-1 +√-1) (a – b√-1)
= (i + i) (a – bi) = 2i(a – bi)
= 2ai – 2bi² = 2ai + 2b

Question 4.
Multiply (x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
Solution:
(x – \(\frac{1+\sqrt{-3}}{2}\)) (x – \(\frac{1-\sqrt{3}}{2}\))
= (x + \(\frac{-1-i \sqrt{3}}{2}\)) (x + \(\frac{-1+i \sqrt{3}}{2}\))
= (x + ω) (x + ω²) = x² + ω²x + ωx + ω³
= x² + x (ω ² + ω) + 1 = x² –  x + 1

Question 5.
Express with rational denominator. \(\frac{1}{3-\sqrt{-2}}\)
Solution:
\(\begin{aligned}
& \frac{1}{3-\sqrt{-2}}=\frac{1}{3-\sqrt{2} \mathrm{i}}=\frac{3+\sqrt{2} \mathrm{i}}{(3-\sqrt{2} \mathrm{i})(3+\sqrt{2} \mathrm{i})} \\
& =\frac{3+\sqrt{2} \mathrm{i}}{9-2 \mathrm{i}^2}=\frac{3+\sqrt{2} \mathrm{i}}{9+2}=\frac{3+\mathrm{i} \sqrt{2}}{11}
\end{aligned}\)

Question 6.
\(\frac{3 \sqrt{-2}+2(-5)}{3 \sqrt{-2}-2 \sqrt{-2}}\)
Solution:

Question 7.
\(\frac{3+2 \sqrt{-1}}{2-5 \sqrt{-1}}+\frac{3-2 \sqrt{-1}}{2+5 \sqrt{-1}}\)
Solution:

Question 8.
\(\frac{a+x \sqrt{-1}}{a-x \sqrt{-1}}-\frac{a-x \sqrt{-1}}{a+x \sqrt{-1}}\)
Solution:

Question 9.
\(\frac{(x+\sqrt{-1})^2}{x-\sqrt{-1}}-\frac{\left(x-\sqrt{-1^2}\right)}{x+\sqrt{-1}}\)
Solution:

Question 10.
\(\frac{(a+\sqrt{-1})^3-(a-\sqrt{-1})^3}{(a+\sqrt{-1})^2-(a-\sqrt{-1})^2}\)
Solution:

Question 11.
Find the value of (- i)⁴ⁿ⁺³; when n is positive.
Solution:
(- i)⁴ⁿ⁺³
= (-i⁴ⁿ) (-i)³ = 1(- i³) = – (-i) = i

Question 12.
Find the square root of (a + 40i) + \(\sqrt{9-40 \sqrt{-i}}\)
Solution:

Question 13.
Express in the form of a + ib:
(i) \(\frac{3+5 i}{2-3 i}\)
Solution:

(ii) \(\frac{\sqrt{3}-i \sqrt{2}}{2 \sqrt{3}-i \sqrt{3}}\)
Solution

(iii) \(\frac{(\mathrm{I}+\mathrm{i})^2}{3-\mathrm{i}}\)
Solution:

(iv) \(\frac{(a+i b)^3}{a-i b}-\frac{(a-i b)^3}{a+i b}\)
Solution:

(v) \(\frac{1+i}{1-i}\)
Solution:
\(\frac{1+i}{1-i}\) = \(\frac{(1+i)^2}{2}=\frac{1-1+2 i}{2}\)
= i = 0 + i

Question 14.
Express the following points geometrically in the Argand plane.
(i) 1
Solution:
1 = 1 + i0 = (1, 0)

(ii) 3i
Solution:
3i = 0 + 3i = (0, 3)

(iii) – 2
Solution:
– 2 = – 2 + i0 = (- 2, 0)

(iv) 3 + 2i
Solution:
3 + 2i = (3, 2)

(v) – 3 + i
Solution:
– 3 + i = (- 3, 1)

(vi) 1-i
Solution:
1 – i = (1, – 1)

Question 15.
Show that the following numbers are equidistant from the origin:
√2 +i, 1 + i√2, i√3
Solution:
|√2 + i| = \(\sqrt{(\sqrt{2})^2+1^2}\) = √3
|1 + i√2| = \(\sqrt{1^2+(\sqrt{2})^2}\) = √3
and |i√3| = √3
∴ The points are equidistant from the origin.

Question 16.
Express each of the above complex numbers in the polar form.
Solution:

Question 17.
If 1, ω, ω² are the three cube roots of unity, prove that (1 + ω²)⁴ = ω
Solution:
L.H.S. = (1 + ω²)⁴ = (- ω)⁴ = ω⁴
= ω³ .ω = 1. ω = ω
= R.H.S. (Proved)

Question 18.
(1 – ω+ ω² ) (1 + ω – ω² ) = 4
Solution:
L.H.S. = (1 – ω+ ω² ) (1 + ω – ω² )
= (- ω – ω )(- ω² – ω² ) (∴ 1 + ω + ω² = 0)
= (- 2ω² – 2ω²) = 4ω³ = 4 = R.H.S.

Question 19.
(1 – ω) (1 – ω)² (1 – ω⁴) (1 – ω⁵) = 9
Solution:
L.H.S. =
(1 – ω) (1 – ω)² (1 – ω⁴) (1 – ω⁵)
= (1 – ω) (1 – ω²) (1 – ω) (1 – ω²)
= (1 – ω)² (1 – ω²)²
= {(1 – ω) (1 – ω²)}²
= (1 – ω² – ω + ω³)²
= {3 – (ω² + ω + 1)}²
= (3)² = 9 = R.H.S.

Question 20.
(2 + 5ω + 2ω² )⁶ = (2 + 2ω + 5ω² )⁶ =729
Solution:
L.H.S. = (2 + 5ω + 2ω² )⁶
(2 + 2ω² + 5ω)⁶ = {2(1 + ω² ) + 5ω}⁶
(- 2ω + 5ω)⁶ = (3ω)⁶ = 729ω⁶ = 729
Again, (2 + 2ω + 5ω² )⁶
= {2(1 + ω) + 5ω² )⁶
= (- 2ω² + 5ω² )⁶ = (3ω²)⁶
= 729ω¹² =729
∴ (2 + 5ω + 2ω² )⁶ = (2 + 2ω + 5ω² )⁶ =729

Question 21.
(1 – ω + ω² ) (1 – ω² + ω⁴) (1 – ω⁴ +ω²)  ….to 2n factors = 2²ⁿ
Solution:
L.H.S. = (1 – ω + ω² ) (1 – ω² + ω⁴) (1 – ω⁴ +ω²)  ….to 2n factors = 2²ⁿ
= (- ω – ω) (1 – ω² + ω) (1 – ω + ω² )  ….to 2n factors = 2²ⁿ
= (- 2ω) (- ω² – ω² ) (- ω – ω)  ….to 2n factors = 2²ⁿ
= [(- 2ω)(- 2ω) … to n factors] × [(- 2ω²)(- 2ω²) …. to n factors]
= (- 2ω)ⁿ × (- 2ω²)ⁿ = (4ω³)ⁿ = 4ⁿ = 2²ⁿ
R.H.S. (Proved)

Question 22.
Prove that x³ + y³ + z³– 3xyz
= (x + y + z) (x + ωy + ω²z) (x + yω² + zω)
Solution:
R.H.S. = (x + y + z) (x + ωy + ω²z) (x + yω² + zω)
= (x + y + z) (x + xyω² + zxω + xyω + y²ω³+ yz ω2 + zxω² + yzω⁴ + z²ω³)
= (x + y + z) [x² + y² + z² + xy (ω² + ω) +yz (ω² + ω) + zx(ω² + ω)]
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S. (Proved)

Question 23.
If x = a + b, y = aω + b ω², z = aω² + bω show that
(1) xyz = a³ + b³
Solution:
L.H.S. = xyz
= (a + b) (a ω + b ω²) (a ω² + b ω)
= (a + b) (a² ω³ + ab ω² + ab ω⁴ + b² ω³)
= (a + b) {a² + b² + ab(ω² + ω)}
= (a + b) (a² – ab + b²) = a³ + b³ = R.H.S. (Proved)

(2) x² + y² + z² = 6ab
Solution:
L.H.S. = x² + y² + z²
= (a + b)² + (a ω + b ω²)² (a ω² + b ω)²
= a² + b² + 2ab + a² ω² + b² ω⁴ + 2ab ω³ + a² ω⁴ + b² ω² + 2ab ω³
= a² + a² ω² + a² ω + b² + b² ω + b² ω² + 2ab + 2ab + 2ab
= a²(1 + ω² + ω) + b² (1 + ω + ω²) + 6ab
= 0 + 0 + 6ab = 6ab = R.H.S. (Proved)

(3) x³ + y³ + z³ = 3(a³ + b³)
Solution:
L.H.S. = x³ + y³ + z³
= (a + b)³ + (a ω + b ω²)³ + (aω² + b²)³
= a³ + 3a²b + 3ab² + b³ + a³ ω³ + 3a²b² bω² + 3a ωb² ω⁴ + b³ ω⁶ + a³ ω⁶ + 3a² ω⁴bω + 3a ω²b² ω² + b³ ω³
= a³ + a³ + a³ + b³ + b³ + b³ + 3a²b(1 + ω⁴ + ω⁵) + 3ab² (1 + ω⁵ + ω⁴)
= 3a³ + 3b³ + 0 + 0 = 3 (a³ + b³) = R.H.S. (Proved)

Question 24.
If ax + by + cz = X, cx + by + az = Y, bx + ay + cz = Z
show that (a² + b² + c² – ab – bc- ca) (x² +y² + z² – xy – yz – zx) = X² + Y² + Z² – YZ – ZX – XY
Solution:
L.H.S.
= (a² + b² + c² – ab – bc – ca) (x² + y² + z² – xy – yz – zx)
= (a + b ω + cω²) (a + bω² + cω) (x + yω + zω)² (x + yω² + z ω) (Refer Q.No.22)
= {(a + bω + cω²) (x + yω + zω²)} {(a + bω² + cω) (x + yω² + zω)}
= (ax + ayω + azω² + bx ω + byω² + bzω³ + cxω² + cyω³ + czω⁴) × (ax + ay ω² + azω + bxω² + byω⁴ + bzω³ + cxω + cyω³ + czω²)
= {(ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω²} x (ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω²}
= (X + Yω² + Zω) (X + Yω + Zω²)
= X² + Y² + Z² – XY – YZ – ZX (Refer Q. No. 22)
(Where X = ax + cy + bz.
Y = cx + by + az.
Z = bx + ay + cz).

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(a)

Question 1.
What is the total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3}?
Solution:
The total number of functions that can be defined from the set {1, 2} to the set {1, 2, 3} is 3² = 9.

Question 2.
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face is thrown twice in succession, what is the total number of outcomes thus obtained?
Solution:
A die of six faces marked with the integers 1, 2, 3, 4, 5, and 6 one on each face, is thrown twice in succession.
∴ The total number of outcomes is 6² = 36.

Question 3.
Five cities A, B, C, D, and E are connected to each other by straight roads. What is the total number of such roads?
Solution:
Five cities A, B, C, D, and E are connected to each other by straight roads.
∴ The number of such roads is \(\frac{{ }^5 \mathrm{P}_2}{2 !}\) = 10

Question 4.
What is the total number of different diagonals of a given pentagon?
Solution:
The total number of different diagonals of a given pentagon is
\(\frac{{ }^5 \mathrm{P}_2}{2 !}\) – 5 = 10 – 5 = 5

Question 5.
There are two routes joining city A to city B and three routes joining B to city C. In how many ways can a person perform a journey from A to C?
Solution:
There are two routes joining city A to city B and three routes joining B to city C.
∴ By the fundamental principle of counting, the total number of journeys in which a person can perform from A to C is 2 × 3 = 6.

Question 6.
How many different four-lettered words can be formed by using the four letters a, b, c, and d, while the letters can be repeated?
Solution:
The number of different words that can be formed by using the four letters a, b, c, and d, while the letters can be repeated is 4⁴ = 256.

Question 7.
What is the sum of all three-digit numbers formed by using the digits 1, 2, and 3?
Solution:
The 3-digit numbers that can be formed by using the digits 1, 2, and 3 are 123, 132, 213, 231, 312, and 321.
∴ Their sum = 1332.

Question 8.
How many different words with two letters can be formed by using the letters of the word JUNGLE, each containing one vowel and one consonant?
Solution:
The word ‘Jungle’ contains 2 vowels and 4 consonants. Each word contains one vowel and one consonant.
The number of different words formed.
2 × ⁴P₁ × ²P₁ = 2 × 4 × 2 = 16

Question 9.
There are four doors leading to the inside of a cinema hall. In how many ways can a person enter into it and come out?
Solution:
There are four doors leading to the inside of a cinema hall. A person can enter into it and come out in 4² = 16 different ways. (By the principle of counting.)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 5 Principles Of Mathematical Induction Ex 5 Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 5 Principles Of Mathematical Induction Exercise 5

Prove the following by induction.

Question 1.
1 + 2 + 3 + …… + n = \(\frac{n(n+1)}{2}\)
Solution:
Let p_n be the given statement

Question 2.
1² + 2² + …… + n² = \(\frac{n(n+1)(2 n+1)}{6}\)
Solution:
Let p_n be the given statement
when n = 1
1² =1 = \(\frac{1(1+1)(2 \times 1+1)}{6}\)
P₁ is true
Let P_k be true.
i.e. 1² + 2² + … + k² = \(\frac{k(k+1)(2 k+1)}{6}\)
we shall prove P_k + 1 is true i.e., 1² + 2² + … + k² + (k + 1)²
\(=\frac{(k+1)(k+2)(2 k+3)}{6}\)

∴ P_k+1 is true
∴ P_n is true of all values of n∈N.

Question 3.
1 + r + r²+ …. + rⁿ = \(\frac{r^{n+1}-1}{r-1}\)
Solution:
Let P_n be the given statement

Question 4.
5ⁿ – 1 is divisible by 4.
Solution:
Let P_n = 5ⁿ – 1
When n = 1,
5¹ – 1 = 4 is divisible by 4.
∴ P₁, is true.
Let P_k be true i.e.,
5^k – 1 is divisible by 4.
Let 5^k+1 – 1 = 4m, me Z
Now 5k + 1 – 1 = 5^k. 5 – 5 + 4
= 5 (5^k – 1) + 4
= 5 × 4m + 4 = 4 (5m + 1)
which is divisible by 4.
∴ P_k+1 is true.
∴ P_n is true for all values of n ∈ N

Question 5.
7²ⁿ + 2^3n-3  3^n-1 is divisible by 25 for any natural number n > 1.
Solution:
Let 7²ⁿ + 2^3n-3 . 3^n-1
when n = 1, 7¹ + 2⁰ . 3⁰ ⇒ 49 + 1 = 50
which is divisible by 25.
∴ P₁ is true. Let P_k be true.
i.e., 72^k + 2^3k-3 3^k-1 is divisible by 35
Now \(7^{2 \overline{k+1}}+2^{3 \overline{k+1}-3} 3^{\overline{k+1}-1}\)
=7^2k+2 + 2^3k. 3^k
= 7^2k. 7² + 2^3k-3 2³. 3^k-1. 3¹
= 7^2k. 49 + 2^3k-3. 3^k-1. 24
= 7^2k (25 + 24) + 24. 2^3k-3. 3^k-1
= 7^2k. 25 + 24 (7^2k + 2^3k-3. 3^k-1)
= 7^2k. 25 + 24 × 25m
Which is divisible by 25 (∵ P_k is true)
∴ P_k+1 is true
∴ P_n is true for all values of n > 1.

Question 6.
7. 5^2n-1 + 23ⁿ⁺¹ is divisible by 17 for every natural number n ≥ 1.
Solution:
Let P_n = 7. 5²ⁿ – 1 + 2³ⁿ⁺¹.
When n = 1, 7.5 + 2⁴ = 35 + 16 = 51
Which is divisible by 17.
P₁, is true.
Let P_k be true i.e., 7.5^2k-1 + 2^3k+1  is divisible by 17.
Let 7.5^2k-1 + 2^3k+1 = 17 m, m ∈ Z
Now, \(7.5^{2 \overline{k+1}-1}+2^{3 \overline{k+1}+1}\)
= 7.5^2k-1 + 2^3k+4
= 7.5^2k-1 . 5² + 2^3k+1 . 2³
= 25. 7. 5^2k-1 + 8. 2^3k+1
= (17 + 8) 7.5^2k-1 + 8. 2^3k+1
= 17. 7. 5^2k-1 + 8 (7. 5^2k-1 + 2^3k+1)
= 17 × 7 × 5^2k-1 + 8 × 17m
Which is divisible by 17.
Hence P_k+1. is true.
∴ P_n is true for all values of n ≥ 1.

Question 7.
4ⁿ⁺¹ + 15n + 14 is divisible by 9 for every natural number n ≥ 0.
Solution:
Let P_n = 4ⁿ⁺¹+ 15 n + 14
when n = 1, 4² + 15 + 14 = 45 is divisible by 9.
∴ P₁ is true. Let P_k be true.
i.e., 4^k+1 + 15k + 14 is divisible by 9.
Now, 4^k+1+1 + 15 (k + 1) + 14
= 4^k+2 + 15k + 29
= 4^k+1. 4 + 60k + 56 – 45k – 27
= 4 (4^k+1 + 15k + 14) – 9 (5k + 3)
Which is divisible by 9.
∴ P_k+1, is true.
∴ P_n is true for all values of n ≥ 0.

Question 8.
3^(2n-1) + 7 is divisible by 9 for every natural number n ≥ 2.
Solution:
Let P_n = 3^2(n-1) + 7
When n = 2. 3² + 7 = 16 is divisible by 8.
∴ P₂ is true.
Let P_k be true.
i.e., 5^2(k-1) + 7 is divisible by 8.
Let 3^2k-2 + 7 = 8m. m ∈ Z.
Now \(3^{2(\overline{k+1}-1)}\) + 7 = 3^2k + 7
= 3^2k-2. 3² + 63 – 56
= 9(3^2k-2 + 7) – 56
= 9 × 8m – 56 = 8 (9m – 7)
Which is divisible by 8.
P_k+1 is true.
P_n is true for all values on n ≥ 2.

Question 9.
5^(2n-4)  – 6n + 32 is divisible by 9 for every natural number n ≥ 5.
Solution:
Let P = 5^2(n-4) – 6n + 32
For n = 5, P₅ = 5² – 6. 5 + 32
= 25 – 30 + 32 = 27
Which is divisible by 9.
Hence P₅ is true.
Let P_k is true.
Let P_k is divisible by 9.
Let P_k = 5^2(k-4) – 6k + 32 = 9., m ≥ Z
5^2k+2 = 576 m + 24k  + 25 … (1)
we shall prove that P_k+1 is true.
Now 5^2(K+1)+2 – 24(k+1) – 25
= 5² (5^2k+2) – 24k – 24 – 25
= 5²[576m + 24k + 25] – 24k – 24 – 25
= 25 × 576 m + 25 × 24k + 25 × 25 – 24k – 24 – 25
= 25 × 576 m + 576 k + 576
= 576 [25 m + k + 1]
which is divisible by 576
∴ P_k+1 is true.
So by the method of induction P_n is true for all n.
i.e., 5²ⁿ⁺²  – 24n – 25 is divisible by 576 for all n ∈ N.
Hence P_k+1 is true.
So by methods of induction P_n is true.
i.e., 5²ⁿ⁺² – 24n – 25 is divisible by 576 for all n.

Question 10.
\(\frac{1}{1.2}+\frac{1}{2.3}+\ldots+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
Solution:
when n = 1,

Question 11.
1.3 + 2.4 + 3.5 + …….. + n(n + 2) = \(\frac{n(n+1)(2 n+7)}{6}\)
Solution:
when n = 1,
we have 1.3 = 3 = \(\frac{3 \times 6}{6}\)
\(=\frac{1 \times 2 \times 9}{6}=\frac{1(1+1)(2 \times 1+7)}{6}\)

Question 12.
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R [Hint : Write xⁿ⁺¹ – yⁿ⁺¹ = x(xⁿ – yⁿ) + yⁿ(x – y)]
Solution:
Let p(n) is
xⁿ – yⁿ = (x – y)(x^n-1 + x^n-2 y + … + xy^n-2 + y^n-1); x, y ∈ R
Step – 1:
For n = 2
x² – y² = (x – y) (x + y) (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., x^k – y^k = (x – y)(x^k-1 + x^k-2y + … +xy^k-2 + y^k-1)
Step – 3:
Let us prove P_k+1 is true.
i.e., x^k+1 – y^k+1 = (x – y) (x^k + x^k-1y + … (xy^k-1 + y^k)
L.H.S. = x^k+1 – y^k+1
= x^k+1 – xy^k + xy^k – y^k+1
= x(x^k – y^k) + y^k (x – y)
= x(x – y)(x^k-1 + x^k-2 y + … + xy^k-2 + y^k-1) + y^k(x – y) [by (1)]
= (x – y) [x^k + x^k-1 y + … + xy^k-2 + xy^k-1 + y^k]
= R.H.S.
∴ P_(k+1) is true.
Step – 4:
By Principle Of Mathematical Induction P(n) is true for all n ∈ N.

Question 13.
1 + 3 + 5 + ……. +(2n – 1) = n²
Solution:
Let P(n) is : 1 + 3 + 5 + ……. +(2n – 1) = n².
Step – 1:
For n = 2
L.H.S. = 1 + 3 = 4 = 2² (R.H.S)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
i.e., 1 + 3 + 5 … + (2k – 1) = k² …(1)
Step – 3:
We will prove that P(k + 1) is true
i.e., we want to prove.
1+ 3 + 5 + … + (2k – 1) + (2k + 1) = (k + 1)²
L.H.S. = 1 + 3 + 5 + … + (2k – 1) + (2k + 1)
= k² + 2k + 1          [By – (1)]
= (k + 1)² = R.H.S.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 1 + 3 + 5 ….+ (2n – 1) = n²

Question 14.
n > n; n is a natural number.
Solution:
Let P(n) is 2ⁿ > n
Step – 1:
2¹ > 1 (True)
∴ P(1) is true.
Step – 2:
Let P(k) is true.
⇒ 2^k > k
Step – 3:
We shall prove that P(k + 1) is true
i.e., 2^k+1 > k + 1
Now 2^k+1 = 2.2k > 2k ≥ k + 1 for k ∈ N.
∴ 2^k+1 > k + 1
⇒ P(k + 1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n.
i.e., 2ⁿ > n for n ∈ N

Question 15.
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Solution:
Let P(n) is
(1, 2, 3 … n)³  > 8 (1³ + 2³ + 3³ + … + n³), for n > 3.
Step – 1:
For n = 4
(1. 2. 3. 4)³ = 24³ = 13824
8(1³+ 2³ + 3³ + 4³) = 808
∴ (1. 2 . 3 . 4)³ > 8(1³ + 2³ + 3³ + 4³)
∴ P(4) is true.
Step- 2:
Let P(k) is true.
(1. 2. 3…….k)³ > 8(1³ + 2³ + 3³ + …… + k³)
Step – 3:
We shall prove that P_(k+1) is true.
i.e., (1. 2. 3. …….. k(k+1))³ > 8(1³ + 2³ + … + k³ + (k + 1)³)
Now (1. 2. 3. …….. k(k + 1)³)
= (1. 2. 3 … k)³ (k + 1)³
> 8 (1³ + 2³ + … k³) (k + 1)³
> 8 (1³ + 2³ + … k³) + 8(k + 1)³
= 8 (1³ + 2³ + … + k³ + (k + 1)³)
P_(k+1) is true.
Step – 4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N and n > 3.

Question 16.
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\) for every positive integer n.
Solution:
Let P(n) is
\(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{3 n+1}>\)



Step-4:
By the Principle of Mathematical Induction P(n) is true for all n ∈ N.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(a)

Question 1.
Find the distance between the following pairs of points.
(i) (3, 4), (-2, 1);
Solution:
Distance between points (3, 4) and (-2, 1) is
\(\sqrt{(3+2)^2+(4-1)^2}=\sqrt{25+9}=\sqrt{34}\)

(ii) (-1, 0), (5, 3)
Solution:
The distance between the points (-1, 0) and (5, 3) is
\(\sqrt{(-1-5)^2+(0-3)^2}\)
= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

Question 2.
If the distance between points (3, a) and (6, 1) is 5, find the value of a.
Solution:
Distance between the points. (3, a) and (6, 1) is
\(\sqrt{(3-6)^2+(a-1)^2}=\sqrt{9+(a-1)^2}\)
∴ \(\sqrt{9+(a-1)^2}=5\)
or, 9 + (a – 1)² = 25
or, (a – 1)² = 16
or, a – 1 = ± 4
a = 1 ± 4 = 5 or, – 3

Question 3.
Find the coordinate of the points which divides the line segment joining the points A (4, 6), B (-3, 1) in the ratio 2: 3 internally. Find also the coordinates of the point which divides \(\overline{\mathbf{A B}}\) in the same ratio externally.
Solution:

Question 4.
Find the coordinates of the mid-point of the following pairs of points.
(i) (-7, 3), (8, -4);
Solution:
Mid-point of the line segment joining the points (-7, 3) and (8, -4) are \(\left(\frac{-7+8}{2}, \frac{3-4}{2}\right)=\left(\frac{1}{2},-\frac{1}{2}\right)\)

(ii) (\(\frac{3}{4}\), -2), (\(\frac{-5}{2}\), 1)
Solution:
Mid-point of the line segment joining the points. (\(\frac{3}{4}\), -2) and (\(\frac{-5}{2}\), 1) is,
\(\left(\frac{\frac{3}{4}-\frac{5}{2}}{2}, \frac{-2+1}{2}\right)=\left(\frac{-7}{8}, \frac{-1}{2}\right)\)

Question 5.
Find the area of the triangle whose vertices are (1, 2), (3, 4) (\(\frac{1}{2}\), \(\frac{1}{4}\))
Solution:
Area of the triangle whose vertices are (1, 2), (3, 4) and (\(\frac{1}{2}\), \(\frac{1}{4}\)) is

Question 6.
If the area of the triangle with vertices (0, 0), (1, 0), (0, a) is 10 units, find the value of a.
Solution:
Area of the triangle with vertices (0, 0),(1,0), (0, a), is \(\frac{1}{2}\) × 1 × a = \(\frac{a}{2}\)
∴ \(\frac{a}{2}\) = 10 or a = 20

Question 7.
Find the value of a so that the points (1, 4), (2, 7), (3, a) are collinear.
Solution:
As points (1, 4), (2, 7), (3, a) are collinear, we have the area of the triangle with vertices (1, 4), (2, 7), and (3, a) is zero.
∴ \(\frac{1}{2}\) {1(7 – a) + 2(a – 4) + 3 (4 – 7)} = 0
or, 7 – a + 2a – 8 + 12 – 21 =0
⇒ a = 10

Question 8.
Find the slope of the lines whose inclinations are given.
(i) 30°
Solution:
The slope of the line whose inclination is 30°.
tan 30° = \(\frac{1}{\sqrt{3}}\)

(ii) 45°
Solution:
Slope = tan 45° = +1

(iii) 60°
Solution:
Slope = tan 60° = √3

(iv) 135°
Solution:
Slope = tan 135° = – 1

Question 9.
Find the inclination of the lines whose slopes are given below.
(i) \(\frac{1}{\sqrt{3}}\)
Solution:
The slope of the line is \(\frac{1}{\sqrt{3}}\)
∴ tan θ = \(\frac{1}{\sqrt{3}}\) or, θ = 30°
∴ The inclination of the line is 30°

(ii) 1
Solution:
Slope = 1 = tan 45°
∴ The inclination of the line is 45°.

(iii) √3
Solution:
Slope = √3 = tan 60°  ∴ θ = 60°
∴ Inclination = 60°

(iv) – 1
Solution:
Slope = – 1 = tan 135°
∴ Inclination = 135°

Question 10.
Find the angle between the pair of lines whose slopes are ;
(i) \(\frac{1}{\sqrt{3}}\), 1
Solution:

(ii) √3, -1
Solution:

Question 11.
(a) Show that the points (0, -1), (-2, 3), (6, 7), and (8, 3) are vertices of a rectangle.
Solution:


∴ The opposite sides are equal and two consecutive sides are perpendicular. So it is a rectangle.

(b) Show that the points (1, 1), (-1, -1), and (-√3, √3) are the vertices of an equilateral triangle.
Solution:

Question 12.
Find the coordinates of the point P(x, y) which is equidistant from (0, 0), (32, 10), and (42, 0).
Solution:

Question 13.
If the points (x, y) are equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Question 14.
The coordinate of the vertices of a triangle are (α₁, β₁), (α₂, β₂), and (α₃, β₃). Prove that the coordinates of its centroid is \(\left(\frac{\alpha_1+\alpha_2+\alpha_3}{3}, \frac{\beta_1+\beta_2+\beta_3}{3}\right)\)
Solution:

Question 15.
Two vertices of a triangle are (0, -4) and  (6, 0). If the medians meet at the point (2, 0), find the coordinates of the third vertex.
Solution:

∴ \(\frac{6+x}{3}\) = 2, \(\frac{-4+y}{3}\) = 0
⇒ x = 0, y = 4
∴ The coordinates of the 3rd vertex are (0, 4).

Question 16.
If the point (0, 4) divides the line segment joining(-4, 10) and (2, 1) internally, find the point which divides it externally in these same ratios.
Solution:

Question 17.
Find the ratios in which the line segment joining (-2, -3) arid (5, 4) is divided by the coordinate axes and hence find the coordinates of these points.
Solution:

Question 18.
In a triangle, one of the vertices is at (2, 5) and the centroid of the triangle is at (-1, 1). Find the coordinates of the midpoint of the side opposite to the given angular point.
Solution:

Question 19.
Find the coordinates of the vertices of a triangle whose sides have midpoints at (2, 1), (-1, 3), and (-2, 5).
Solution:

∴ x₂ + x₃ – 4 or, x₂ – 4 – 3 = 1
∴ x₁ = – 4 – x₂ = -4 – 1 = -5
Similarly y₁ + y₂ + y₃ = 5 + 1 + 3 = 9
As y₁ + y₂ = 10
we have y₃ = 9 – 10 = – 1
Again y₁ + y₃ = 6
or, y₁ = 6 – y₃ = 6 + 1 = 7
and y₂ = 10 – y₁ = 10 – 7 = 3
∴ The coordinates of A, B, and C are (-5, 7), (1, 3), and (3, -1).

Question 20.
If the vertices of a triangle have their coordinates given by rational numbers, prove that the triangle cannot be equilateral.
Solution:
Let us choose the contradiction method. Let the triangle is equilateral if the co¬ ordinate of the vertices is rational numbers.
Let ABC be an equilateral triangle with vertices A (a, 0), B (a, 0), and C (b, c) where a, b, c are rational.

⇒ a² = b² + c² = \(\frac{a^2}{4}\) + c²
⇒ c² =  a² – \(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\) ⇒ c = \(\frac{\sqrt{3}}{2}\) a     ….(2)
Now b = \(\frac{a}{2}\), c = \(\frac{\sqrt{3}}{2}\) a
If a is rational then b is rational but c is irrational, i.e., the coordinates of the vertices are not rational, which contradicts the assumption.
Hence assumption is wrong.
So the triangle cannot be equilateral if the coordinate of the vertices is rational numbers.

Question 21.
Prove that the area of any triangle is equal to four times the area of the triangle formed by joining the midpoints of its sides.
Solution:

∴ The area of triangle ABC is four times the area of triangle DEF. (Proved)

Question 22.
Find the condition that the point (x, y) may lie on the line joining (1, 2) and (5, -3).
Solution:

∴ As points A, B, and C are collinear, we have the area of the triangle ABC as 0.
∴ \(\frac{1}{2}\) {1(-3 – y) + 5(y – 2) + x(2 + 3)} = 0
or, – 3 – y + 5y – 10 + 5x = 0
or, 5x + 4y = 13

Question 23.
Show that the three distinct points (a², a), (b², b), and (c², c) can never be collinear.
Solution:
Area of the triangle with vertices (a², a), (b², b) , and (c², c) is
\(\frac{1}{2}\) {a²(b – c) + b²(c – a) c²(a – b)}
= (a – b)(b – c)(a – c)
which is never equal to zero except when a = b = c, hence the points are not collinear.

Question 24.
If A, B, and C are points (-1, 2), (3, 1), and (-2, -3) respectively, then show that the points which divide BC, CA, and AB in the ratios (1: 3), (4: 3) and (-9: 4) respectively are collinear.
Solution:
Let the points P, Q, and R divides \(\overline{\mathrm{BC}}\), \(\overline{\mathrm{CA}}\), in \(\overline{\mathrm{AB}}\) the ratio 1: 3, 4: 3 and -9: 4

Question 25.
Prove analytically :
(a) The line segment joining the midpoints of two sides of a triangle is parallel to the third and half of its length.

Solution:
Let the coordinates of the triangle ABC be (x₁, y₁), (x₂, y₂) and (x₃, y₃)
The points D and E are the midpoints of the sides \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)

(b) The altitudes of a triangle are concurrent.
Solution:

(c) The perpendicular bisectors of the sides of a triangle are concurrent.
Solution:

(d) An angle in a semicircle is a right angle.
Solution:

∴ The angle in a semicircle is a right angle. (Proved)

Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 10 Sequences and Series Ex 10(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 10 Sequences and Series Exercise 10(a)

Question 1.
Which of the following in a sequence?
(i) f(x) = [x], x ∈ R
(ii) f(x) = |x|, x ∈ R
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N
Solution:
(iii) f(x) = \(\sqrt[n]{\pi}\) ,n∈ N is a sequence f(n) : N → X, X ⊂ R.

Question 2.
Determine if (t_n) is an arithmetic sequence if :
(i) t_n = an² + bn
Solution:
t_n = an² + bn
⇒ t_n+1 = a(n + 1)² + b(n – 1)
⇒ t_n+1 – t_n = a{(n + 1)² – n²} + b{n + 1 – n}
= a(2n + 1) + b
Which is not independent of n.
∴ (t_n) is not an A.P.

(ii) t_n = an + b
Solution:
t_n = an + b
⇒ t_n+1 = a(n + 1) + b
Now t_n+1 – t_n
= {a(n + 1) + b} – {an + b}
= a (constant)
∴ (t_n) is an arithmetic sequence.

(iii) t_n = an² + b
Solution:
t_n = an² + b
⇒ t_n+1 = a(n + 1)² + b
∴ t_n+1 – t_n = a[(n + 1)² – n²] + b – b
= a(2n + 1)
(does not independent of n)
∴ (t_n) is not an arithmetic sequence.

Question 3.
If a geometric series converges which of the following is true about its common ratio r?
(i) r > 1
(ii) -1 < r < 1
(iii) r > 0
Solution:
(ii) -1 < r < 1

Question 4.
If an arithmetic series ∑tn converges, which of the following is true about t_n?
(i) t_n < 1
(ii) |t_n| < 1
(iii) t_n = 0
(iv) t_n → 0
Solution:
(iii) t_n = 0

Question 5.
Which of the following is an arithmetic-geometric series?
(i) 1 + 3x + 7x² + 15x³+ ….
(ii) x + \(\frac{1}{2}\)x + \(\frac{1}{3}\)x² + ….
(iii) x + (1 + 2)x² + (1 + 2 + 3)x³ +…
(iv) x + 3x² + 5x³ + 7x⁴ + …
Solution:
(iv) x + 3x² + 5x³ + 7x⁴ + … is an arithmetic geometric series with a = 1, d = 2, r = x.

Question 6.
For an arithmetic sequence (t_n) t_p = q, t_q = p, (p ≠ q), find t_n.
Solution:
t_p = q ⇒ a + (p – 1)d = q    ……(1)
t_q = p ⇒ a + (q – 1)d = p    ……(2)
From (1) and (2) we have (p – q)d = q – p
⇒ d = (-1)
Putting d = (-1) in (1)
we have a = p + q – 1
∴ t_n = a + (n – 1)d
= (P + q – 1) + (n – 1) (-1)
= p + q – n

Question 7.
For an arithmetic series, ∑a_n S_p = q and S_q = p (p ≠ q) find S_p+q
Solution:
S_p = q and S_q = p

Question 8.
The sum of a geometric series is 3. The series of squares of its terms have a sum of 18. Find the series.
Solution:

Question 9.
The sum of a geometric series is 14, and the series of cubes of its terms have a sum of 392 find the series.
Solution:


∴ The series is \(\sum_{n=1}^{\infty} \frac{7}{2^{n-1}}\)

Question 10.
Find the sum as directed
(i) 1 + 2a + 3a² + 4a³ + …..(first n terms(a ≠ 1))
Solution:

(ii) 1 + (1 + x)y + (1 + x + x²)y² + …..(to infinity)
Solution:
Let S_∞ = 1 + (1 + x)y + (1x + x²)y² + …
⇒ S_n = 1 + (1 + 1 + x)y + (1 + x + x²)y² + ……+(1 + x + …. x^n-1)y^n-1

(iii) 1 + \(\frac{3}{5}\) + \(\frac{7}{25}\) + \(\frac{15}{125}\) + \(\frac{31}{625}\) + …..(to infinity)
Solution:

(iv) 1 + 4x + 8x² + 13x³ + 19x⁴ + …..(to infinity). Assuming that the series has a sum for |x| < 1.
Solution:

(v) 3.2 + 5.2² + 7.2³ + …..(first n terms)
Solution:
Sn = 3.2 + 5.22 + 7.3 + ….n terms = 2[3 + 5.2 + 7.22 + ….n terms]

Question 11.
Find the sum of the infinite series.
(i) \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots\)
Solution:

= 1 – \(\frac{1}{n+1}\)
∴ \(S_{\infty}=\lim _{n \rightarrow \infty} S_n=1\)

(ii) \(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)
Solution:
\(\frac{1}{1 \cdot 2 \cdot 3}+\frac{1}{2 \cdot 3 \cdot 4}+\frac{1}{3 \cdot 4 \cdot 5}+\ldots\)

(iii) \(\frac{1}{2 \cdot 5 \cdot 8}+\frac{1}{5 \cdot 8 \cdot 11}+\frac{1}{8 \cdot 11 \cdot 14}+\ldots\)
Solution:
Here t_n = \(\frac{1}{(3 n-1)(3 n+2)(3 n+5)}\)
The denominator of t_n is the product of 3 consecutive terms of A.P. Now multiplying and dividing by (3n + 5) – (3n – 1) we have
\(t_n=\frac{(3 n+5)-(3 n-1)}{6(3 n-1)(3 n+2)(3 n+5)}\)

(iv) \(\frac{3}{1^2 \cdot 2^2}+\frac{5}{2^2 \cdot 3^2}+\frac{7}{3^2 \cdot 4^2}+\ldots\) [Hint : take t_n = \(\frac{2 n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{n^2(n+1)^2}\)]
Solution:

(v) \(\frac{1}{1 \cdot 5}+\frac{1}{3 \cdot 7}+\frac{1}{5 \cdot 9}+\ldots .\)
Solution:

Question 12.
Find S_n for the series.
(i) 1.2 + 2.3 + 3.4 + ….
Solution:

(ii) 1.2.3 + 2.3.4 + 3.4.5 + …
Solution:

(iii) 2.5.8 + 5.8.11 + 8.11.14 +…
Solution:

(iv) 1.2.3.4 + 2.3.4.5 + 3.4.5.6 + …
[Hint : t_n = (3n – 1) (3n + 2)(3n + 5)]
Solution:

(v) 1.5 + 2.6 + 3.7 + …
[Hint: t_n = n(n + 4) is not a product of two successive terms of an A.P. for the term following n should be n+1, not n+4. So the method of previous exercises is not applicable. Instead, write t_n = n² + 4n and find S_n = \(\sum_{k=1}^n k^2+4 \sum_{k=1}^n k\) applying formulae]
Solution:

(vi) 2.3 + 3.6 + 4.11 + …
[Hint : Take t_n = (n + 1) (n² + 2)]
Solution:

(vii) 1.3² + 2.5² + 3.7² + ….
Solution:

Question 13.
Find the sum of the first n terms of the series:
(i) 5 + 6 + 8 + 12 + 20 + …
Solution:

(ii) 4 + 5 + 8 + 13 + 20 + …
Solution:

Question 14.
(i) Find the sum of the product of 1,2,3….20 taken two at a time. [Hint: Required sum = \(\frac{1}{2}\left\{\left(\sum_{k=1}^{20} k\right)^2-\sum_{k=1}^{20} k^2\right\}\)]
Solution:
We know that
(x₁ + x₂ + x₃ + …. + x_n)²
= (x²₁ + xⁿ₂ + … + x²_n) + 2 (Sum of all possible Products taken two at a time)
∴ 2 (Sum of products of 1. 2, 3,…… 20 taken two at a time)
= (1 + 2 + 3 + … 20)² – (1² + 2² + … + 20²)
\(=\left(\frac{20 \times 21}{2}\right)^2-\frac{20(20+1)(40+1)}{6}\)
= (210)² – 70 × 41
= 44100 – 2870 = 41230
∴ The required sum = \(\frac{41230}{2}\) = 20615

(ii) Do the same for 1, 3, 5, 7,….19.
Solution:
Here the required sum

Question 15.
If a = 1 + x + x² + ….. and b = 1 + y + y² + ….|x| <  1 and |y| <  1, then prove that 1 – xy + x²y² + x³y³ + ….. =  \(\frac{a b}{a+b-1}\)
Solution:

Question 16.
If a, b, c are respectively the pth, qth, rth terms of an A.P., then prove that a(q – r) + b(r – p) + c(p – q) = 0
Solution:
Let the first term of an A.P. = A and the common difference = D.
According to the question
t_P = a, t_q = b, t_r = c
⇒ A + (p – q)D = a      …..(1)
A + (q – 1)D = b           …..(2)
A + (r – 1)D = c            …..(3)
L.H.S = a(q – r) + b (r – p) + c (p – q)
= (A + (p – 1)D) (q – r) + (A + (q – 1)D)
(r- p) + (A + (r – 1)D) (p – q)
= A (q – r + r – p + p – q) + D [(p – 1)
(q – r) + (q – 1) (r – p) + (r – 1) (p – q)]
= D (pq – pr + qr – pq + pr – qr) – D (q – r + r – p + p – q) = 0

Question 17.
If \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in A.P. and a + b + c ≠ 0, prove that \(\frac{\mathbf{b}+\mathbf{c}}{\mathbf{a}}, \frac{\mathbf{c}+\mathbf{a}}{\mathbf{b}}, \frac{\mathbf{a}+\mathbf{b}}{\mathbf{c}}\) are in A.P.
Solution:

Question 18.
If a², b², c² are in A.P. Prove that \(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\) are in A.P.
Solution:

Question 19.
If \(\frac{b+c}{a}, \frac{c+a}{b}, \frac{\mathbf{a}+\mathbf{b}}{c}\) are in A.P.,prove that \(\frac{\mathbf{1}}{\mathbf{a}}, \frac{\mathbf{1}}{\mathbf{b}}, \frac{\mathbf{1}}{\mathbf{c}}\) are inA.P.given a + b + c ≠ 0
Solution:

Question 20.
If (b – c)², (c – a)², (a – b)² are in A.P., prove that \(\frac{1}{b-c}, \frac{1}{c-a}, \frac{1}{a-b}\) are in A.P.
Solution:

Question 21.
If a, b, c are respectively the sum of p, q, r terms of an A.P., prove that \(\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)\) = 0
Solution:

Question 22.
If a, b, c,d are in G.P., prove that (a² + b² + c²)(b² + c² + d²) = (ab + bc + cd)².
Solution:
Let a, b, c, and d are in G.P.
Let the common ratio = r
⇒ b = ar, c = ar², d = ar³
LHS = (a² + b² + c²) (b² + c² + d²)
= (a² + a²r² + a²r⁴) (a²r² + a²r⁴ + a²r⁶)
= a⁴r² (1 + r² + r⁴)²
= (a²r + a²r³ + a²r⁵)²
= (a.ar + ar.ar² + ar².ar³)²
= (ab + bc + cd)² = R.H.S. (proved)

 

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 8 Permutations and Combinations Ex 8(c) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 8 Permutations and Combinations Exercise 8(c)

Question 1.
Compute the following :
(i) ¹²C₃
Solution:
¹²C₃ = \(\frac{(12) !}{3 ! 9 !}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2}\) = 220

(ii) ¹⁵C₁₂
Solution:
¹⁵C₁₂ = \(\frac{(15) !}{(12) ! 3 !}=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2}\)
= 5.7.13 = 455

(iii) ⁹C₄ + ⁹C₅
Solution:
⁹C₄ + ⁹C₅ = \(\frac{9 !}{4 ! 5 !}+\frac{9 !}{5 ! 4 !}\)
\(=\frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}\) × 2 = 252

(iv) ⁷C₃ + ⁶C₄ + ⁶C₃
Solution:
⁷C₃ + ⁶C₄ + ⁶C₃ = ⁷C₃ + ⁶C₄ + ⁶C4-1
= ⁷C₃ + ⁶⁺¹C₄ = ⁷C₃ + ⁷C₄
(∴ ⁿc_r + ⁿC_r-1– = ⁿ⁺¹c_r)
= ⁷C₄ + ⁷C_4-1 = ⁷⁺¹C₄
= ⁸C₄ = \(\frac{8 !}{4 !(8-4) !}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}\) = 70

(v) ⁸C₀ + ⁸C₁ + …….. + ⁸c₈
Solution:
⁸C₀ + ⁸C₁ + …….. + ⁸c₈ = 2⁸ = 256

Question 2.
Solve :
(i) ⁿC₄ = ⁿC₁₁ ;
Solution:
ⁿC₄ = ⁿC₁₁ ;  (∴ n = 4 + 11 = 15)

(ii) ²ⁿC₃ : ⁿC₃ = 44: 5
Solution:
²ⁿC₃ : ⁿC₃ = \(\frac{44}{5}\)
⇒ \(\frac{2 n !}{2 n-3 !} / \frac{n !}{n-3 !}=\frac{44}{5}\)

Question 3.
Find n and r if ⁿP_r = 1680, ⁿC_r = 70.
Solution:
ⁿP_r = 1680, ⁿC_r = 70
∴ \(\frac{{ }^n \mathrm{P}_r}{{ }^n \mathrm{C}_r}=\frac{1680}{70}\)
or, r ! = 24 = 4!
∴ r = 4
Again, ⁿC_r = 70 or ⁿC₄ = 70
or, \(\frac{n !}{4 !(n-4) !}=70\)
or, n(n – 1) (n – 2) (n – 3)
= 70 × 4! = 7 × 10 × 4 × 3 × 2
= 8 × 7 × 6 × 5
or, n(n – 1) (n – 2) (n – 3)
or, 8(8 – 1) (8 – 2) (8 – 3)
∴ n = 8

Question 4.
How many diagonals can an n-gon(a polygon with n sides) have?
Solution:
A polygon of n – sides has n vertices.
∴ The number of st. lines joining the n-vertices is ⁿC₂.
∴ The number of diagonals is ⁿC₂ – n

Question 5.
If a set A has n elements and another set B has m elements, what is the number of relations from A to B?
Solution:
If |A| = n, |B| = m
then |A × B| = mn
∴ The number of possible subsets of
A × B = 2^mn
∴ The number of relations from A to B is 2^mn.

Question 6.
From five consonants and four vowels, how many words consist of three consonants and two vowels?
Solution:
Words of consisting of 3 consonants and 2 vowels are to be formed from five consonants and 4 vowels.
∴ The number of ways = ⁵C₃ × ⁴C₂
Again, 5 letters can be arranged among themselves in 5! ways.
∴ The total number of ways
= ⁵C₃ × ⁴C₂ × 5! = 10 × 6 × 120 = 7200.

Question 7.
In how many ways can a committee of four gentlemen and three ladies be formed out of seven gentlemen and six ladies?
Solution:
A committee of 4 gentlemen and 3 ladies is to be formed out of 7 gentlemen and 6 ladies.
∴ The number of ways in which the committee can be formed.
⁷C₄ × ⁶C₃ = \(\frac{7 \cdot 6 \cdot 5}{3.2} \times \frac{6 \cdot 5 \cdot 4}{3 \cdot 2}\) = 700

Question 8.
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily. In how many ways can this be done? Find also the number of ways such that at least 3 black balls can be drawn.
Solution:
A bag contains 4 black and 5 white balls out of which 6 balls are drawn arbitrarily.
∴ The number of ways in which balls are drawn \({ }^9 \mathrm{C}_6=\frac{9 \cdot 8 \cdot 7}{3 \cdot 2 \cdot 1}\) = 84 as the total number of balls is 9. If at least 3 black balls are drawn, then the drawing can be made as follows.

Black(4)	White(5)
3	3
4	2

The number of ways in which at least 3 black balls are drawn
= (⁴C₃ × ⁵C₃) + (⁴C₄ × ⁵C₂)
= (4 × 10) + (1 × 10) = 50

Question 9.
How many triangles can be drawn by joining the vertices of a decagon?
Solution:
A decagon has 10 vertices and 3 noncollinear points are required to be a triangle.
∴ The number of triangles formed by the joining of the vertices of a decagon is
¹⁰C₃ = \(\frac{10 !}{3 ! 7 !}=\frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1}\) = 120

Question 9.
How many triangles can be drawn by joining the vertices and the center of a regular hexagon?
Solution:
A regular hexagon has six vertices. Triangles are to be formed by joining the vertices and center of the hexagon. So there is a total of 7 points. So the number of triangles formed.
⁷C₃ = \(\frac{7 !}{3 ! 4 !}=\frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}\) = 35
As the hexagon has 3 main diagonals, which pass through the center hence can not form 3 triangles.
∴ The required number of triangles 35 – 3 = 32

Question 11.
Sixty points lie on a plane, out of which no three points are collinear. How many straight lines can be formed by joining pairs of points?
Solution:
Sixty points lie on a plane, out of which no. 3 points are collinear. A straight line required two points. The number of straight lines formed by joining 60 points is
⁶⁰C₂ = \(\frac{60 !}{2 \times 58 !}=\frac{60 \times 59}{2}\) = 1770

Question 12.
In how many ways can 10 boys and 10 girls sit in a row so that no two boys sit together?
Solution:
10 boys and 10 girls sit in a row so that no two boys sit together. So a boy is to be seated between two girls or at the two ends of the row. So the boys are to be sitted in 11 positions in ¹¹C₁₀ ways. Again 10 boys and 10 girls can be arranged among themselves in 10! and 10! ways respectively.
∴ The total number of ways = ¹¹C₁₀ × 10! × 10! = (11)! × (10)!

Question 13.
In how many ways can six men and seven girls sit in a row so that the girls always sit together?
Solution:
Six men and seven girls sit in a row so that the girls always sit together. Considering the 7 girls as one person, there are a total of 7 persons who can sit in 7! ways. Again the 7 girls can be arranged among themselves in 7! ways.
∴ The total number of arrangements
= 7! × 7!
= (7!)²

Question 14.
How many factors does 1155 have that are divisible by 3?
Solution:

∴ In order to be a factor of 1155 divisible by 3, we have to choose one or two of 5, 7, and 11 along with 3 or 3 alone.
∴ The number of ways = ³C₁ + ³C₂ + ³C₀ = 2³ – 1 = 7
∴ The number of factors is 7 excluding 1155 itself.

Question 15.
How many factors does 210 have?
Solution:

∴ We can choose at least one 2, 3, 5, or 7  to be a factor of 210.
∴ The number of factors.
= ⁴C₁ + ⁴C₂ + ⁴C₃ + ⁴C₄ = 2⁴ – 1 = 15
∴ The number of factors is 210 is 15. (Including 215 itself and excluding 1).

Question 16.
If n is a product of k distinct primes what is the total number of factors of n?
Solution:
n is a product of k distinct primes.
∴ In order to be a factor, of n, we have chosen at least one of k distinct primes.
∴ The number of ways = ^kC₁ + ^kC₂ + ……… ^kC_k-1 = 2^k  – 1 – 1
∴ The number of factors of n is 2^k – 2.
(Excluding 1 as 1 is not prime. It is also not include n.)

Question 17.
If m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn, what is the total number of factors of m (excluding 1)?
Solution:
m has the prime factor decomposition P₁^r1, P₂^r2 ….. P_n^rn,
∴ m = P₁^r1, P₂^r2 ….. P_n^rn,
P₁ is a factor of m which occurs r₁ times. Each of the factors P₁^r1 will give rise to (r₁ + 1) factors.
Similarly
P₂^r2 gives (r₂ + 1) factors and so on.
∴ The total number of factors (r₁ + 1) (r₂ + 1) ….. (r_n + 1) – 1 (including m).

Question 18.
If 20! were multiplied out, how many consecutive zeros would it have on the right?
Solution:
If 20! were multiplied out, then the number of consecutive zeros on the right is 4. due to the presence of 4 x 5, 10, 14 x 15,20.

Question 19.
How many factors of 10,000 end with a 5 on the right?
Answer:
We have 1000 = 2⁴ × 5⁴ The factors of 10000 ending with 5 are 5, 5 × 5 = 25, 5 × 5 × 5 = 125
5 × 5 × 5 × 5 = 625
∴ There are 4 factors ending With 5.

Question 20.
A man has 6 friends. In how many ways can he invite two or more to a dinner party?
Solution:
A man has 6 friends. He can invite 2 or more of his friends to a dinner party.
∴ He can invite 2, 3, 4, 5, or 6 of his friends in
⁶C₂ + ⁶C₃ + ⁶C₄ + ⁶C₅ + ⁶C₆ = 2⁶ – ⁶C₀ – ⁶C₁
= 64 – 7 = 57 ways.

Question 21.
In how many ways can a student choose 5 courses out of 9 if 2 courses are compulsory?
Solution:
A student is to choose 5 courses out of 9 in which 2 courses are compulsory. as 2 courses are compulsory, he is to choose 3 courses out of 7 courses in ⁷C₃ = 35 ways.

Question 22.
In how many ways can a student choose five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together?
Solution:
A student chooses five courses out of the courses? C₁, C₂, …………. C₉ if C₁, C₂ are compulsory and C₆, C₈ cannot be taken together.
∴ He is to choose 3 courses out of C₃, C₄, ………… ,C₈, C₉.
Without taking any restrictions 3 courses out of C₃, C₄, …… C₉, i.e. from 7 courses in ⁷C₃ ways. If C₆, C₈ are taken together then one course only to be choosen from C₃, C₄, C₅, C₇, C₉ by ⁵C₁ ways. Hence required number of ways.
= ⁷C₃ – ⁵C₁
= \(\frac{7 \times 6 \times 5}{3 \times 2 \times 1}\) – 5
= 35 – 5 = 30 ways.

Question 23.
A cricket team consisting of 11 players is to be chosen from 8 batsmen and 5 bowlers. In how many ways can the team be chosen so as to include at least 3 bowlers?
Solution:
A cricket team consisting of 1 1 player is to be chosen from 8 batsmen and 5 bowlers, as to include at least 3 bowlers.
The selection can be made as follows :

Batsmen(8)	Bowlers(5)
8	3
7	4
6	5

The number of selections is
(⁸C₈ × ⁵C₃) + (⁸C₇ × ⁵C₄) + (⁸C₆ × ⁵C₅)
= (1 × 10) + (8 × 5) + (28 × 1)
= 10 + 40 + 28 = 78

Question 24.
There are n + r points on a plane out of which n points lie on a straight line L and out of the remaining r points that lie outside L, no three points are collinear. What is the number of straight lines that can be formed by joining pairs of their points?
Solution:
There are n + r points on a plane out of which n points are collinear and out of which r points are not collinear.
∴ We can form a straight line by joining any two points.
n-collinear points form one line and r-non-collinear points form ^rC₂ lines.
Again, each of the r non-collinear points when joined to each of the noncollinear points, forms n lines.
∴ The number of such limes is r x n.
∴ The total number of lines

Question 25.
There are 10 books in a shelf with different titles; five of these have red covers and others have green covers. In how many ways can these be arranged so that the red books are placed together?
Solution:
There are 10 books in a shelf with different titles, 5 of these are red covers and others are green covers considering 5 red-covered books as one book, we have a total of 6 books which can be arranged in 6! ways. The five red cover books are arranged among themselves in 5! ways.
∴ The total number of arrangements
= 5! x 6!

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 11 Straight Lines Ex 11(b) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 11 Straight Lines Exercise 11(b)

Question 1.
Fill in the blanks in each of the following, using the answers given against each of them :
(a) The slope and x-intercept of the line 3x – y + k = 0 are equal if k = _________ . (0, -1, 3, -9)
Solution:
-9

(b) The lines 2x – 3y + 1 = 0 and 3x + ky – 1=0 are perpendicular to each other if k = ___________ . (2, 3, -2, -3)
Solution:
2

(c) The lines 3x + ky – 4 = 0 and k – Ay – 3x = 0 are coincident if k = _____________. (1, -4, 4, -1)
Solution:
4

(d) The distance between the lines 3x – 1 = 0 and x + 3 = 0 is _________ units. (4, 2, \(\frac{8}{3}\), \(\frac{10}{3}\))
Solution:
\(\frac{10}{3}\)

(e) The angle between the lines x = 2 and x – √3y + 1 = 0 is _________. (30°, 60°, 120°, 150°)
Solution:
60°

Question 2.
State with reasons which of the following are true or false :
(a) The equation x = k represents a line parallel to x – axis for all real values of k.
Solution:
False. As the line x = k is parallel to y- axis for all values of k.

(b) The line, y + x + 1 = 0 makes an angle 45° with y – axis.
Solution:
y + x + 1 = 0
∴ Its slope = -1 = tan 135°
∴ It makes 45° with y – axis, as it makes 135° with x – axis. (True)

(c) The lines represented by 2x – 3y + 1 = 0 and 3x + 2y – k = 0 are perpendicular to each other for positive values of k only.
Solution:
2x – 3y + 1 = 0, 3x + 2y – k = 0
∴ \(m_1 m_2=\frac{2}{3} \times \frac{(-3)}{2}=-1\)
∴ The lines are perpendicular to each other for + ve values of k only. (False)

(d) The lines represented by px + 2y – 1 = 0 and 3x + py + 1 = 0 are not coincident for any value of ‘p’.
Solution:
px + 2y – 1 = 0, 3x + py + 1=0
∴ \(\frac{p}{3}=\frac{2}{p}=\frac{-1}{1} \Rightarrow p^2=6\)
and p = -3 or -2
There is no particular value of p for which \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) (True)

(e) The equation of the line whose x and y – intercepts are 1 and -1 respectively is x – y + 1 = 0.
Solution:
Equation of the line whose intercepts 1 and -1 is \(\frac{x}{1}+\frac{y}{-1}\) = 1
or, x – y = 1 (False)

(f) The point (-1, 2) lies on the line 2x + 3y – 4 = 0.
Solution:
Putting x = – 1, y = 2
we have 2 (- 1) + 3 × 2 – 4
= -2 + 6 – 4 = 0
∴ The point (-1, 2) lies on the line 2x + 3 – 4 = 0 (True)

(g) The equation of a line through (1, 1) and (-2, -2) is y = – 2x.
Solution:
The equation of the line through (1, 1) and (-2, -2) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
or, y – 1 = \(\frac{-2-1}{-2-1}\) (x – 1)
or, y – 1 = x – 1
or, x – y = 0 (False)

(h) The line through (1, 2) perpendicular to y = x is y + x – 2 =0.
Solution:
The slope of the line y = x is 1.
∴ The slope of the line perpendicular to the above line is -1.
∴ The equation of the line through (1, 2) having slope – 1 is y – y₁ = m(x – x₁)
or, y – 2 = -1 (x – 1)
or, y – 2= -x + 1
or, x + y = 3 (False)

(i) The lines \(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1 are intersecting but not perpendicular to each other.
Solution:
\(\frac{x}{a}+\frac{y}{b}\) = 1 and \(\frac{y}{a}-\frac{x}{b}\) = 1
∴ \(m_1 m_2=\frac{\left(-\frac{1}{a}\right)}{\frac{1}{b}} \times \frac{\left(-\frac{1}{b}\right)}{\left(-\frac{1}{a}\right)}=-1\)
∴ The lines intersect and are perpendicular to each other. (False)

(j) The points (1, 2) and (3, – 2) are on the opposite sides of the line 2x + y = 1.
Solution:
2x + y = 1
Putting x = 1, y = 2,
we have 2 × 1 + 2 = 4 > 1
Putting x – 3, y = -2,
we have 2 × 3 – 2 – 4 > 1
∴ Points (1, 2) and (3, – 2) lie on the same side of the line 2x + y = 1 (False)

Question 3.
A point P (x, y) is such that its distance from the fixed point (α, 0) is equal to its distance from the y – axis. Prove that the equation of the locus is given by, y² = α (2x – α).
Solution:

Question 4.
Find the locus of the point P (x, y) such that the area of the triangle PAB is 5, where A is the point (1, -1) and B is the tie point (5, 2).
Solution:

= \(\frac{1}{2}\) (-3x + 4y + 7) = 5
or, – 3x + 4y + 7 = 10
or, 3x – 4y + 3 =0 which is the locus of the point P (x, y).

Question 5.
A point is such that its distance from the point (3, 0) is twice its distance from the point (-3, 0). Find the equation of the locus.

Question 6.
Obtain the equation of straight lines:
(a) Passing through (1, – 1) and making an angle 150°.
Solution:
The slope of the line
= tan 150° = –\(\frac{1}{\sqrt{3}}\)
∴ The equation of the line is y – y₁ = m(x – x₁)
or, y + 1 = –\(\frac{1}{\sqrt{3}}\) (x – 1)
or, y√3 + √3 = -x + 1
or, x + y√3 + √3 – 1 = 0

(b) Passing through (-1, 2) and making intercept 2 on the y-axis.
Solution:
Let the equation of the line be
y – mx + c or, y = mx + 2
∴ As the line passes through (-1, 2)
we have 2 = – m + 2, or, m = 0
∴ Equation of the line is y = 2.

(c) Passing through the points (2, 3) and (-4, 1).
Solution:
The equation of the line is

(d) Passing through (- 2, 3) and a sum of whose intercepts in 2.
Solution:
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}\) = 1 where a + b = 2     …….(1)
Again, as the line passes through the point (-2, 3), we have \(\frac{-2}{a}+\frac{3}{b}\) = 1      ………(2)
From (1), we have a= 2 – b
∴ From (2) \(\rightarrow \frac{-2}{2-b}+\frac{3}{b}=1\)
or, – 2b + 6 – 3b = (2 – b)b
or, 6 – 5b = 2b – b²
or, b² – 7b + 6 = 0
or, (b – 6)(b – 1) = 0
∴ b = 6, 1
∴ a =2 – b = 2 – 6 = -4
or, 2 – 1 = 1
∴ Equation of the lines are \(\frac{x}{-4}+\frac{y}{6}\) = 1 or \(\frac{x}{1}+\frac{y}{1}\) = 1
i, e. -3x + 2y = 12 or, x + y = 1

(e) Whose perpendicular distance from the origin is 2 such that the perpendicular from the origin has indication 150°.
Solution:
Here p = 2, α = 150°
The equation of the line in normal form is x cos α + y sin α = p
or, x cos 150° + y sin 150° = 2
or, \(\frac{-x \sqrt{3}}{2}+y \cdot \frac{1}{2}\) = 2
or, -x √3 + y = 4
or, x√3 – y + 4 = 0

(f) Bisecting the line segment joining (3, – 4) and (1, 2) at right angles.
Solution:
The slope of the line \(\overline{\mathrm{AB}}\) is

(g) Bisecting the line segment joining, (a, 0) and (0, b) at right angles.
Solution:
Refer to (f)

(h) Bisecting the line segments joining (a, b), (a’, b’) and (-a, b), (a’, -b’).
Solution:

(i) Passing through the origin and the points of trisection of the portion of the line 3x + y – 12 = 0 intercepted between the coordinate axes.
Solution:

(j) Passing through (-4, 2) and parallel to the line 4x – 3y = 10.
Solution:
Slope of the line 4x – 3y = 10 is \(\frac{-4}{-3}=\frac{4}{3}\)
∴ The slope of the line parallel to the above line is \(\frac{4}{3}\).
∴ Equation of the line through (- 4, 2) and having slope \(\frac{4}{3}\) is y – y₁ = m(x – x₁)
or, y – 2 = \(\frac{4}{3}\) (x + 4)
or, 3y – 6 = 4x + 16
or, 4x – 3y + 22 = 0

(k) Passing through the point (a cos³ θ, a sin³ θ) and perpendicular to the straight line x sec θ + y cosec θ = α.
Solution:
The slope of the line x sec θ + y cosec θ = a is \(\frac{-\sec \theta}{{cosec} \theta}\) = -tanθ
∴ Slope of the required line  = cot θ
∴ Equation of the line through (a cos³ θ, a sin³ θ) is y – y₁ = m(x – x₁)
or, y – a sin³ θ = cot θ(x – a cos³ θ)
or, y – a sin³ θ = \(\frac{\cos \theta}{\sin \theta}\) (x – a cos³ θ)
or y sin θ – a sin⁴ θ = x cos  θ – a cos⁴ θ
or (x cos θ – y sin θ) + a(sin⁴ θ – cos⁴ θ) = 0

(l) Which passes through the point (3, -4) and is such that its portion between the axes is divided at this point internally in the ratio 2: 3.
Solution:

(m) which passes through the point (α, β) and is such that the given point bisects its portion between the coordinate axis.
Solution:

x = 2α , y = 2β
∴ Equation of the line \(\overleftrightarrow{\mathrm{AB}}\) is \(\frac{x}{2 \alpha}+\frac{y}{2 \beta}\) = 1(Intercept form)

Question 7.
(a) Find the equation of the lines that is parallel to the line 3x + 4y + 7 = 0 and is at a distance 2 from it.
Solution:
3x + 4y + 7 = 0
or, \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) = 0(Normal form)
∴ Equation of the lines parallel to the above line and 2 units away from it are \(\frac{3 x}{5}+\frac{4 y}{5}+\frac{7}{5}\) ± 2 = 0
or, 3x + 4y + 7 ± 10 = 0
∴ 3x + 4y + 17 = 0 and 3x + 4y – 3 = 0

(b) Find the equations of diagonals of the parallelogram formed by the lines ax + by = 0, ax + by + c = 0, lx + my = 0, and lx + my + n = 0. What is the condition that this will be a rhombus?
Solution:

(c) Find the equation of the line passing through the intersection of 2x – y – 1 = 0 and 3x – 4y + 6 = 0 and parallel to the line x + y – 2 = 0.
Solution:
Let the equation of the required line be (2x – y – 1) + λ(3x – 4y + 6) = 0
or, x(2 + 3λ) + λ(-1 – Aλ) + 6λ – 1 = 0
As this line is parallel to the line x + y – 2 = 0
we have their slopes are equal.
∴ \(-\left(\frac{2+3 \lambda}{-1-4 \lambda}\right)=\frac{-1}{1}\)
or, 2 + 3λ = -1 – 4λ
or, 7λ = -3 or, λ = \(\frac{-3}{7}\)
∴ Equation of the line is (2x – y – 1) – \(\frac{3}{7}\) (3x – 4y + 6) = 0
or, 14x – 7y – 1 – 9x + 12y – 18 = 0
or, 5x + 5y – 25= 0
or, x + y = 5

(d) Find the equation of the line passing through the point of intersection of lines x + 3y + 2 = 0 and x – 2y – 4 = 0 and perpendicular to the line 2y + 5x – 9 = 0.
Solution:
Let the equation of the line be (x + 3y + 2) + λ(x – 2y – 4) = 0
or, x(1 + λ) + y(3 – 2λ) + 2 – 4λ = 0
As this line is perpendicular to the line 2y + 5x – 9 = 0.
We have the product of their slopes is -1.
∴ \(\frac{1+\lambda}{3-2 \lambda} \times \frac{5}{2}\) = -1
or, 5 + 5λ = – 6 + 4λ
or, λ = -6 – 5 = -11.
∴ Equation of the required line is (x + 3y + 2) – 11(x – 2y – 4) = 0
or, x + 3y + 2- 11x + 22y + 44 = 0
or, – 10x + 25y + 46 = 0
or, 10x – 25y – 46 = 0

(e) Find the equation of the line passing through the intersection of the lines x + 3y – 1 = 0 and 3x – y + 1 = 0 and the centroid of the triangle whose vertices are the points (3, -1) (1, 3) and (2, 4).
Solution:
Let the equation of the required line (x + 3y – 1) + λ(3x – y + 1) = 0   … (1)
Again, the centroid of the triangle with vertices (3, – 1), (1, 3), and (2, 4) is \(\left(\frac{3+1+2}{3}, \frac{-1+3+4}{3}\right)\) = (2, 2)
As line (1) passes through (2, 2), we have (2 + 6 – 1) +1(6 – 2 + 1) = 0
or, 7 + 5λ = 0 or, λ = \(\frac{-7}{5}\)
∴ Equation of the line (x + 3y – 1) – \(\frac{7}{5}\) (3x – y + 1) = 0
or, 5x + 15y – 5 – 21x + 7y – 7 = 0
or, 22y – 16x – 12 = 0
or, 11y – 8x – 6 = 0
or, 8x – 11y + 6 = 0

Question 8.
If lx + my + 3 = 0 and 3x – 2y – 1 = 0 represent the same line, find the values of l and m.
Solution:
lx + my + 3 = 0 and 3x – 2y – 1 = 0 represents the same line
∴ \(\frac{l}{3}=\frac{m}{-2}=\frac{3}{-1}\)
∴ l = -9, m = 6

Question 9.
Find the equation of sides of a triangle whose vertices are at (1, 2), (2, 3), and (-3, -5).
Solution:
Equation of \(\overline{\mathrm{AB}}\) is \(y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\)
\(y-2=\frac{3-2}{2-1}(x-1)\)
or, y – 2 = x – 1
or, x – y + 1 = 0

Question 10.
Show that origin is within the triangle whose sides are given by equations, 3x – 2y = 1, 5x + 3y + 11 = 0, and x – 7y + 25 = 0.
Solution:



∴ The origin lies within the triangle ABC.

Question 11.
(a) Find the equations of straight lines passing through the point (3, -2) and making an angle 45° with the line 6x + 5y = 1.

or, 11x – y = 35, x + 11y + 19 = 0

(b) Two straight lines are drawn through the point (3, 4) inclined at an angle 45° to the line x – y – 2 = 0. Find their equations and obtain area included by the above three lines.
Solution:

Slope of L₂ = 0
Then as L₂ ⊥ L₃
Slope of L₃ = ∞
∴ Equation of L₂ is
y – y₁ =m(x – x₁)
or, y – 4 = 0(x – 3) = 0
or, y = 4
∴ Equation of L₃ is y – 4 = ∞ (x – 3)
or, x – 3 = 0 or, x = 3
∴ Sloving L₁ and L₂, we have
x – y – 2 = 0, y = 4
or, x = 6
The coordinates of A are (6, 4).
Again solving L1 and L3, we have
x – y – 2 = 0, x = 3
or, y = x – 2 = 3 – 2 = 1
∴ The coordinates of B are (3, 1).
Area of the triangle PAB is

(c) Show that the area of the triangle formed by the lines given by the equations y = m₁x + c₁,y = m₂x + c_2, and x = 0 is \(\frac{1}{2} \frac{\left(c_1-c_2\right)^2}{\left[m_2-m_1\right]}\)
Solution:

Question 12.
Find the equation of lines passing through the origin and perpendicular to the lines 3x + 2y – 5 = 0 and 4x + 3y = 7. Obtain the coordinates of the points where these perpendiculars meet the given lines. Prove that the equation of a line passing through these two points is 23x + 11y – 35 = 0.
Solution:
The slopes of the line 3x + 2y – 5 = 0 and 4x + 3y = 7 are \(\frac{-3}{2}\) and \(\frac{-4}{3}\)
∴ Slopes of the lines perpendicular to the above lines are \(\frac{2}{3}\) and \(\frac{3}{4}\)
∴ Equation of the lines through the origin and having slopes \(\frac{2}{3}\) and \(\frac{3}{4}\)
y = \(\frac{2x}{3}\) and y = \(\frac{3x}{4}\)
Now solving 3x + 2y – 5 = 0 and y = \(\frac{2x}{3}\)
we have 3x + \(\frac{4x}{3}\) – 5 = 0
or, 9x + 4x – 15 = 0
or, x = \(\frac{15}{13}\)
∴ y = \(\frac{2 x}{3}=\frac{2}{3} \times \frac{15}{13}=\frac{10}{13}\)
∴ The perpendicular y = \(\frac{2 x}{3}\) meets the line 3x + 2y – 5 = 0 at \(\left(\frac{15}{13}, \frac{10}{13}\right)\)
Again, solving 4x + 3y = 7 and y = \(\frac{3 x}{4}\)
we have 4x + 3 × \(\frac{3 x}{4}\) = 7
or, 16x + 9x = 28 or, x = \(\frac{28}{25}\)

Question 13.
(a) Find the length of a perpendicular drawn from the point (-3, -4) to the straight line whose equation is 12x – 5y + 65 = 0.
Solution:
The length of the perpendicular drawn from the point (- 3, -4) to the straight line 12x – 5y + 65 = 0 is

(b) Find the perpendicular distances of the point (2, 1) from the parallel lines 3x – 4y + 4 = 0 and 4y – 3x + 5 = 0. Hence find the distance between them.
Solution:
The distance of the point (2, 1) from the line 3x – 4y + 4 = 0 is \(\left|\frac{3 \times 2-4 \times 1+4}{\sqrt{9+16}}\right|=\frac{6}{5}\)
Again distance of the point (2, 1) from the line 4y – 3x + 5 = 0 is

(c) Find the distance of the point (3, 2) from, the line x + 3y – 1 = 0, measured parallel to the line 3x – 4y + 1 = 0
Solution:

Let the coordinates of M be (h, k).
As \(\overline{\mathrm{PM}} \| \mathrm{L}_1\), We have their slopes are equal.

(d) Find the distance of the point (-1, -2) from the line x + 3y – 7 = 0, measured parallel to the line 3x + 2y – 5 = 0
Solution:
Slope of the line 3x + 2y – 5 = 0 is \(\left(-\frac{3}{2}\right)\)
Equation of the line through (-1, -2) and parallel to this line is y + 2 = – \(\frac{3}{2}\) (x + 1)
⇒ 2y + 4 = -3x – 3
⇒ 3x + 2y + 7 = 0 …(1)
Given line is : x + 3y – 7 = 0    ….(2)
from (1) and (2) we get 7y – 28 = 0 Py = 4 and x = \(-\frac{35}{7}\) = -5
Thus the required distance is \(\sqrt{(-1+5)^2+(-2-4)^2} \quad=\sqrt{16+36}\)
= √52 = 2√3 units.

(e) Fine the distance of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) from the origin.
Solution:
The equation of the line passing through the points (a cos α, a sin α) and (a cos β, a sin β) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) x – x₁
or, y – a sin α

Question 14.
Find the length of perpendiculars drawn from the origin on the sides of the triangle whose vertices are A( 2, 1), B (3, 2), and C (- 1, -1).
Solution:

Question 15.
Show that the product of perpendicular from the points \(\left(\pm \sqrt{a^2-b^2}, 0\right)\) upon the straight line \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 is b²
Solution:

Question 16.
Show that the lengths of perpendiculars drawn from any point of the straight line 2x + 11y – 5 = 0 on the lines 24x + 7y – 20 = 0 and 4x – 3y – 2 = 0 are equal to each other.
Solution:
Let P(h, k) is any point on the line
2k + 11y – 5 = 0
∴ 2h + 11k – 5 = 0
Now the length of the perpendicular from P on the line 24x + 7y – 20 = 0 is

Clearly d₁ = d₂

Question 17.
If p and p’ are the length of perpendiculars drawn from the origin upon the lines x sec α + y cosec α = 0 and x cos α – y sin α – a cos 2α = 0
Prove that 4p² + p’² = a²
Solution:

Question 18.
Obtain the equation of the lines passing through the foot of the perpendicular from (h, k) on the line Ax + By + C = 0 and bisect the angle between the perpendicular and the given line.
Solution:

Slope of the line L is \(\frac{-\mathrm{A}}{\mathrm{B}}\)
∴ Slope of the line \(\overline{\mathrm{PM}} \text { is } \frac{\mathrm{B}}{\mathrm{A}}\)
∴ Equation of the line \(\overline{\mathrm{PM}}\) is y – y₁ = m(x – x₁)
or, y – k = \(\frac{\mathrm{B}}{\mathrm{A}}\) (x – h)
or, Ay – Ak =Bx – Bh
or, Bx – Ay + Ak – Bh = 0
∴ Equation of the bisectors of the angles between the lines L and \(\overline{\mathrm{PM}}\) is
\(\frac{\mathrm{A} x+\mathrm{B} y+\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\pm \frac{\mathrm{B} x-\mathrm{A} y+\mathrm{A} k-\mathrm{B} h}{\sqrt{\mathrm{B}^2+\mathrm{A}^2}}\)
or, Ax + By + C = ± (Bx – Ay +Ak – Bh)

Question 19.
Find the direction in which a straight line must be drawn through the point(1, 2) such that its point of intersection with the line x + y – 4 = 0 is at a distant \(\frac{1}{3} \sqrt{6}\) from this point.
Solution:

Question 20.
A triangle has its three sides formed by the lines x + y = 3, x + 3y = 3, and 3x + 2y = 6. Without solving for the vertices, find the equation of its altitudes and also calculate the angles of the triangle.
Solution:

or 1 + 3λ = – 3 – 6λ
or 9λ = – 4 or , λ = \(\frac{-4}{9}\)
∴ Equation of \(\overline{\mathrm{AD}}\) is (x + y – 3) – \(\frac{4}{9}\) (3x + 2y – 6) = 0
or, 9x + 9y – 27 – 12x – 8y + 24 = 0
or, -3x + y – 3 = 0
or, 3x – y + 3 = 0
Let the equation of \(\overline{\mathrm{BE}}\) be (x + y – 3) + 1(x + 3y – 3) = 0
or, x(1 + λ) + y( 1 + 3λ) – 3 – 3λ = 0
As \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\)
we have \(\frac{1+\lambda}{1+3 \lambda} \times \frac{3}{2}\) = -1
or, 3 + 3λ = -2 – 6λ
or, 9λ = – 5 or λ = \(\frac{-5}{9}\)
∴ Equation of \(\overline{\mathrm{BE}}\) is (x + y – 3) – \(\frac{5}{9}\) (x + 3y – 3) = 0
or 9x + 9y – 27 – 5x – 15y + 15 = 0
or, 4x – 6y – 12 = 0
or, 2x – 3y – 6 = 0
Let the equation of \(\overline{\mathrm{CE}}\) be (3x + 2y – 6) + λ (x + 3y – 3) = 0
x(3 + λ) + y (x + 3λ) – 6 – 3λ = 0
As \(\overline{\mathrm{CF}} \perp \overline{\mathrm{AB}} .\)
we have \(\frac{3+\lambda}{2+3 \lambda}\) × 1 = -1
or, 3 + λ = -2 – 3λ
or, 4λ = -2 – 3 = -5
or, λ = \(\frac{-5}{4}\)
∴ Equation of is \(\overline{\mathrm{CF}}\) (3x + 2y – 6) \(\frac{-5}{4}\) (x + 3y – 3) = 0
or, 12x + 8y – 24 – 5x – 15y + 15 =0
or, 7x – 7y – 9 = 0

Question 21.
A triangle has its vertices at P(1, -1), Q(3, 4) and R(2, 5). Find the equation of altitudes through P and Q and obtain the coordinates of their point of intersection. (This point is called the ortho-center of the triangle.)
Solution:

Question 22.
(a) Show that the line passing through (6, 0) and (-2, -4) is concurrent with the lines
2x – 3y – 11 = 0 and 3x – 4y = 16
Solution:
The equation of the line through (6,0) and (-2, -4) is

(b) Show that the lines lx + my + n = 0 mx + ny + 1 = 0 and nx + ly + m = 0 are concurrent, l + m + n = 0
Solution:
As the lines
lx + my + n = 0
mx + ny + 1 = 0
and nx + ly + m = 0 are concurrent.

Question 23.
Obtain the equation of the bisector of the acute angle between the pair of lines.
(a) x + 2y = 1, 2x + y + 3 = 0
Solution:
Equation ofthe bisectors ofthe angles between the lines x + 2y – 1 = 0 and 2x + y + 3 = 0 are \(\frac{x+2 y-1}{\sqrt{1^2+2^2}}=\pm \frac{2 x+y+3}{\sqrt{2^2+1^2}}\)
or, x + 2y – 1 = ± (2x + y + 3)
∴ x + 2y – 1 = 2x + y + 3 and
x + 2y – 1= -2x – y – 3
∴ x – y + 4 = 0 and 3x + 3y + 2 = 0
Let θ be the angle between x + 2y – 1 = 0 and x – y + 4 = 0
∴ tan θ = \(\frac{a_1 b_2-a_2 b_1}{a_1 a_2+b_1 b_2}\)
\(=\frac{1 \cdot(-1)-(+1) \cdot 2}{1 \cdot 1+2(-1)}\)
\(=\frac{-1-2}{1-2}=\frac{-3}{-1}=3\)
sec² θ = 1 + tan² θ = 1 + 9 = 10
cos² θ = 1/10
cos θ = \(\frac{1}{\sqrt{10}}<\frac{1}{\sqrt{2}}\) ⇒ θ > 45°
∴ x – y + 4 = 0 is the obtuse angle bisector.
⇒ 3x + 3y + 2 = 0 is acute angle bisector.

(b) 3x – 4y = 5, 12y – 5x = 2
Solution:
Given equation of lines are
3x – 4y – 5 = 0    …..(1) and  5x – 12y + 2 = 0     ……(2)
Equation of bisectors of angles between these Unes are:

Question 24.
(a) Find the coordinates of the center of the inscribed circle of the triangle formed by the line x cos α + y sin α = p with the coordinate axes.
Solution:


\(\left(\frac{p}{\sin \alpha+\cos \alpha+1}, \frac{p}{\sin \alpha+\cos \alpha+1}\right)\)

(b) Find the coordinates of the circumcentre and incentre of the triangle formed by the lines 3x – y = 5, x + 2y = 4, and 5x + 3y + 1 = 0.
Solution:

Question 25.
The vertices B, and C of a triangle ABC lie on the lines 3y = 4x and y = 0 respectively, and the side \(\overline{\mathbf{B C}}\) passes through the point (2/3, 2/3). If ABOC is a rhombus, where O is the origin, find the equation of \(\overline{\mathbf{B C}}\) and also the coordinates of A.
Answer:
Let the coordinates of C be (a, 0) so that the length of the side of the rhombus is ‘a’

Question 26.
Find the equation of the lines represented by the following equations.
(a) 4x² – y² = 0
Solution:
4x² – y² = 0
or, (2x + y)(2x – y) = 0
∴ 2x + y = 0 and 2x – y = 0 are the two separate lines.

(b) 2x² – 5xy – 3y²
Solution:
2x² – 5xy – 3y²
or, 2x² – 6xy + xy – 3y² = 0
or, 2x(x – 3y) + y(x – 3y) = 0
or, (x – 3y)(2x + y) = 0
∴ x – 3y = 0 and 2x + y = 0 are the two separate lines.

(c) x² + 2xy sec θ + y² = 0
Solution:
x² + 2xy sec θ + y² = 0
∴ a = 1, b = 2y sec θ, c = y²
x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{2 y \sec \theta \pm \sqrt{4 y^2 \sec ^2 \theta-4 y^2}}{2}\)
= \(\frac{-2 y \sec \theta \pm 2 y \tan \theta}{2}\)
= y (- sec θ ± tan θ)
∴ x = y (- sec θ + tan θ) and x – y (- sec θ – tan θ) are the two separate lines.

(d) 3x² + 4xy = 0
Solution:
3x² + 4xy = 0
or x (3x + 4y) = 0
∴ x = 0 and 3x + 4y = 0 are the two separate lines.

Question 27.
From the equations which represent the following pair of lines.
(a) y = mx; y = nx
Solution:
y – mx = 0, y – nx = 0
or (y – mx) (y – nx) = 0
or, y² – nxy – mxy + mnx² = 0
or, y² – xy (m + n) + mnc² = 0 which is the equation of a pair of lines.

(b) y – 3x = 0 ; y + 3x = 0
Solution:
y – 3x = 0, y + 3x = 0
∴ (y – 3x) (y + 3x) = 0
or, y² – 9x² = 0 which is the equation of a pair of lines.

(c) 2x – 3y + 1 = 0 ; 2x + 3y + 1 = 0
Solution:
2x – 3y + 1 = 0; 2x + 3y + 1 =0
or, (2x – 3y + 1)(2x + 3y + 1) = 0
or, (2x + 1)² – 9y² = 0
or, 4x² + 1 + 4x – 9y² = 0
or, 4x² – 9y² + 4x + 1= 0 which represents a pair of lines.

(d) x = y. x + 2y + 5 = 0
Solution:
x = y, x + 2y + 5 = 0
∴ (x – y) (x + 2y + 5) = 0
or, x² + 2xy + 5x – xy – 2y² – 5y =0
or, x² – 2y² + xy + 5x – 5y = 0 which represents a pair of lines.

Question 28.
Which of the following equations represents a pair of lines?
(a) 2x² – 6y² + 3x +  y + 1 = 0
Solution:
a = 2, b = -6, 2g = 3
2f = 1, c = 1
∴ g = \(\frac{3}{2}\), f = \(\frac{1}{2}\), h = 0
∴ abc + 2fgh – ah² – bg² – ch²
= 2(-6). 1 + 2 × \(\frac{1}{2}\) × \(\frac{3}{2}\) – 0 – (-6) × \(\frac{9}{4}\) – 1 × 0
= -12 + \(\frac{3}{2}\) + \(\frac{27}{2}\) = \(\frac{6}{2}\) = 3 ≠ 0
∴ The given equation does not represent a pair or lines.

(b) 10x² – xy – 6y² – x + 5y – 1 = 0
Solution:
a = 10. 2h = 1
B = -6, 2g = -1
2f = 5. C= -1
∴ h = –\(\frac{1}{2}\), g = –\(\frac{1}{2}\) , f = \(\frac{5}{2}\)
∴ abc + 2fgh – ah² – bg² – ch²
= 10(-6)(-1) + 2 × \(\frac{5}{2}\) × (-\(\frac{1}{2}\)) × (-\(\frac{5}{2}\)) – 10 × \(\frac{2.5}{4}\) – (-6)\(\frac{1}{4}\) – (-1)\(\frac{1}{4}\)
= 60 + \(\frac{5}{4}\) – \(\frac{250}{4}\) + \(\frac{6}{4}\) + \(\frac{1}{4}\)
= \(\frac{240+5+6-250+1}{4}=\frac{2}{4}\)
∴ The given equation does not represent a pair of lines.

(c) xy + x + y + 1 = 0
Solution:
xy + x + y + 1= 0
or, x(y + 1) + 1(y + 1 ) = 0
or (y + 1 )(x + 1) =0
∴ x + 1 = 0
and y + 1 = 0 are the two separate lines,
∴ The given equation represents a pair of lines.

Question 29.
For what value of λ do the following equations represent pair of straight lines?
(a) λx² + 5xy – 2y² – 8x + 5y – λ = 0
Solution:
λx² + 5xy – 2y² – 8x + 5y – λ = 0
∴ a = λ, 2h = 5, b = -2, 2g = -8
2f = 5, c = -1
∴ h = \(\frac{5}{2}\), g = -4, f = \(\frac{5}{2}\)
As the given equation represent a pair of lines, we have abc + 2fgh – ah² – bg² – ch² = 0
or, λ(-2)(-λ) + 2. \(\frac{5}{2}\) (-4). \(\frac{5}{2}\) -λ × \(\frac{25}{4}\) – (-2) (-4)² – (-λ) × \(\frac{25}{4}\) = 0
or, 2λ² – 50 – \(\frac{25 λ }{4}\) + 32 + \(\frac{25 λ }{4}\) = 0
or, 2λ² = 18 or, λ² = 9
λ = ±3

(b) x² – 4xy – y² +6x + 8y + λ = 0
Solution:
Here a = 1, 2h = -1, b = -1, 2g = 6, 2f = 8, c = τ
As the given equation represent a pair of lines, we have
abc + 2fgh – af² – bg² – ch² = 0
⇒ (-1) τ + 2.4.3 (-2) – 1. 4² – (-1). ³2 – τ(-2)² = 0
⇒ -τ – 48 – 16 + 9 – 4τ = 0
⇒ -5τ – 55 = 0 ⇒ τ = -11

Question 30.
(a) Obtain the value of λ for which the pair of straight lines represented by 3x² – 8xy + λy² = 0 are perpendicular to each other.
Solution:
3x² – 8xy + λy² = 0
∴ a = 3. 2h = -8, b = λ
As the pair of lines are perpendicular to each other, we have a + b = 0.
or, 3 + λ = 0 – or, λ = -3

(b) Prove that a pair of lines through the origin perpendicular to the pair of lines represented by px² – 2qxy + ry² = 0 is given by rx² – 2qxy + py² = 0
Solution:
px² – 2qxy + ry² = 0
∴ a = p, b = 2qy, c = ry²

(c) Obtain the condition that a line of the pair of lines ax² + 2hxy + by² = 0,
(i) Coincides with
Solution:

(ii) is perpendicular to, a line of the pair of lines px² + 2qxy + ry² = 0
Solution:

Question 31.
Find the acute angle between the pair of lines given by :
(a) x² + 2xy – 4y² = 0
Solution:
x² + 2xy – 4y² = 0
∴ a=1, 2h = 2, b = -4
∴ tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}=\frac{\pm 2 \sqrt{1+4}}{1-4}\)
\(=\pm \frac{2 \sqrt{5}}{-3}=\mp \frac{2 \sqrt{5}}{3}\)
∴ The acute angle between the pair of lines is tan^-1 \(\frac{2 \sqrt{5}}{3}\)

(b) 2x² + xy – 3y² + 3x + 2y + 1 = 0
Solution:
2x² + xy – 3y² + 3x + 2y + 1 = 0
∴ a = 2, 2h = 1, b = -3, 2g = 3 2f= 2, c = 1.
tan θ = \(\frac{\pm 2 \sqrt{h^2-a b}}{a+b}\)
\(=\pm \frac{2 \sqrt{\frac{1}{4}+6}}{2-3}=\pm \frac{2 \times 5}{2(-1)}\) = ± 5
∴ The acute angle is tan^-1 5

(c) x² + xy – 6y² – x – 8y – 2 = 0
Solution:
Given Equation is x² + xy – 6y² – x – 8y – 2 = 0
here a = 1, 2h = 1, b = -6 thus if 0 is the acute angle between two lines then
tan θ = \(=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
= \(\left|\frac{2 \times 5}{-10}\right|\) = 1
∴ θ = 45°

Question: 32.
Write down the equation of the pair of bisectors of the following pair of lines :
(a) x² – y² = 0 ;
Solution:
x² – y² = 0
∴ a = 1, b = -1, h = 0
∴ The equation of the bisectors of the angles between the pair of lines are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{1+1}=\frac{x y}{0}\)
or, xy = 0

(b) 4x² – xy – 3y² = 0
Solution:
4x² – xy – 3y² = 0
∴ a = 4, 2h = -1, b = -3
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{(a-b)}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{7}=\frac{x y}{\left(-\frac{1}{2}\right)}\)
or, x² – y² = -14xy
or, x² + 14xy – y² = 0

(c) x² cos θ + 2xy – y² sin θ = 0
Solution:
x² cos θ + 2xy – y² sin θ = 0
∴ a = cos θ, 2h = 2, b = – sin θ
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{a-b}=\frac{x y}{h}\)
or, \(\frac{x^2-y^2}{\cos \theta+\sin \theta}=\frac{x y}{1}\)
or, x² – y² = xy(cos θ + sin θ)

(d) x² – 2xy tan θ – y² = 0
Solution:
x² – 2xy tan θ – y² = 0
∴ a = 1, 2h = -2 tan θ, b = -1
∴ Equation of the pair of bisectors are \(\frac{x^2-y^2}{2}=\frac{x y}{-\tan \theta}\)
or, x² – y² = 2xy cot θ
or, x² + 2xy cot θ – y² = 0

Question 33.
If the pair of lines represented by x² – 2pxy – y² = 0 and x² – 2qxy – y² = 0 be such that each pair bisects the angle between the other pair, then prove that pq = -1.
Solution:

Question 34.
Transform the equation: x² + y² – 2x – 4y + 1 = 0 by shifting the origin to (1, 2) and keeping the axes parallel.
Solution:
x² + y² – 2x – 4y + 1 = 0     ……(1)
Let h = 1, k = 2
Taking x’ + h and y = y’ + k we have
(x’ + h)² + (y’ + k)² – 2(x’ + j) -4(y’ + k) + 1=0
or, (x + 1)² + (y’ + 2)² – 2(x’ + 1) – 4(y’+ 2) + 1=0
or, x^‘2 + 1 + 2x’ + y^‘2 + 4 – 4y’ – 2x’ – 2 – 4y’- 8 + 1 = 0
or, x^‘2 + y’² – 4 = 0
∴ The transformed equation is x² + y² = 4

Question 35.
Transform the equation: 2x² + 3y² + 4xy – 12x – 14y + 20 = 0. When referred to parallel axes through(2, 1).
Solution:
2x² + 3y² + 4xy – 12x – 14y + 20 = 0
Let h = 2, k = 1
Taking x = x’ + 1 and y = y’ + 1
we have
2(x’ + 2)² + 3(y’ + 1)² + 4(x’ + k)(y’ + 1) – 12 (x’ + 2)- 14 (y’ + 1) + 20 = 0
or, 2x^‘2 + 8 + 8x’ + 3 + 6y’ + 3y^’2 + 4x’y’ + 4x’ + 8y’ + 8 – 12x’ – 14y’ – 18 = 0
or, 2x^‘2 + 3y^’2 + 4x’y’ + 1=0
The transformed equation is
2x² + 3y² + 4xy + 1 = 0

Question 36.
Find the measure of rotation so that the equation x² – xy + y² = 5 when transformed does not contain xy- term.
Solution:
x² – xy + y² = 5
Taking x = x’ cos α – y’ sin α
y = x’ sin α – y’ cos α
We get (x’ cos α – y’ sin α)² – (x’ cos α – y’ cos α) (x’ sin α + y’ cos α) + (x’ sin α + y’ cos α)² = 5
⇒ x^‘2 cos² α + y^‘2 sin² α – 2x’y sin α.
cos α – x^‘2 sin α. cos α – x’y cos² α + x’y’ sin² α + y^‘2 sin α. cos α + x^‘2 sin² α + y^‘2 cos² α + 2x’y’ sin α cos α = 5
Given that the transformed equation does not xy term.
Hence the co-efficient of x’y’ is zero.
That is sin² α – cos² α = 0
⇒ sin² α = cos² α
⇒ tan² α = 1 ⇒ tan α = 1 ⇒ α= 45°

Question 37.
What does the equation x + 2y – 10 =0 become when the origin is changed to (4, 3)?
Solution:
x + 2y – 10 = 0
Let h = 4, k = 3
Taking x = x’ + 4, y = y’ + 3
we have x’ + 4 + 2 (y’ + 3) – 10 = 0
or, x + 2y’ = 0
∴ The transformed equation is x + 2y = 0.

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(a)

Question 1.
\(\lim _{x \rightarrow 3}\)(x + 4)
Solution:
Clearly, if we take x very close to 3, x + 4 will go very close to 7.
Now let us use ε – δ technique to confirm the result.
Given ε > 0, we seek for δ > 0 depending on ε such that
|x – 3| < δ ⇒ |(x + 4) – 7|< ε
Now |(x + 4) – 7| < ε
if |x – 3| < ε
∴ We can choose ε = 8
Hence for given ε > 0, there exist 8 = ε > 0
such that |x – 3| < δ ⇒ |(x + 4) – 7| < ε
∴ \(\lim _{x \rightarrow 3}\)(x + 4) = 7

Question 2.
\(\lim _{x \rightarrow 1}\)(4x – 1)
Solution:
By taking very close to 1 we have 4x- 1 tends to 3.
Let us use ε – δ technique to confirm the result.
Given ε > 0. We shall find δ > 0 depending on ε such that
|x – 1| < 5 ⇒ |(4x – 1) – 3| < ε
Now |(4x – 1 ) – 3| < ε
if |4x – 1| < ε i.e.|x – 1| < \(\frac{\varepsilon}{4}\)
Let us choose δ = \(\frac{\varepsilon}{4}\)
∴ For given ε > 0 there exists δ = \(\frac{\varepsilon}{4}\) > 0
such that |x – 1| < δ
⇒ |(4x – 1) – 3| < ε
∴ \(\lim _{x \rightarrow 1}\)(4x – 1) = 3

Question 3.
\(\lim _{x \rightarrow 1}\)(√x + 3)
Solution:
As x → 1 we see √x + 3 → 4
We will confirm the result using ε – δ technique
Let ε > 0, we will choose δ > 4
such that |x – 1| < 8 ⇒ |√x + 3 – 4| < ε
Now |√x + 3 – 4| = |√x – 1|
\(=\frac{|x-1|}{|\sqrt{x}+1|}\)
But |√x + 1| > 1
⇒ \(\frac{1}{|\sqrt{x}+1|}\) < 1
⇒ \(\frac{|x-1|}{|\sqrt{x}+1|}<\frac{\delta}{1}\)
∴ (√x + 3) – 4 < \(\frac{\delta}{1}\)
We can take δ < ε i.e. δ = min {1, ε}
∴ |x – 1| < δ ⇒ |(√x + 3) – 4| < ε
for given ε > 0 and (δ = ε)
⇒ \(\lim _{x \rightarrow 1}\)(√x + 3) = 4

Question 4.
\(\lim _{x \rightarrow 0}\) (x² + 3)
Solution:
As x → 0 we observe that x³ + 3 → 3
Let us use ε – δ technique to confirm the result.
Let ε > 0, we seek for a δ > 0 such that
|x – 0| < ε ⇒ |x² + 3 – 3| < ε
Let |x| < 8
Now |x² + 3 – 3| < ε
We have |x|² < ε ⇒| x| < √ε
(∴ |x| and ε are positive.)
∴ we can choose δ = √ε
∴ We have for given δ > 0 there exists
δ = √ε > 0 such that |x| < δ ⇒ |x² + 3 – 3| < ε
∴ \(\lim _{x \rightarrow 0}\) (x² + 3) = 3

Question 5.
\(\lim _{x \rightarrow 0}\) 7
Solution:
If x → 0 we observe that 7 → 7.
Let us use e- 8 technique to confirm the limit.
Let f(x) = 7
Given ε > 0, we will choose a δ > 0
such that |x – 0| < δ ⇒ |f(x) – 7| < ε
Now |f(x) – 7| < ε
If f(x) ∈ (7 – ε . 7 + ε)
But for every x, f(x) = 7
⇒ for|x| < δ also f(x) = 7 ∈ (7 – ε . 7 + ε)
∴ Choosing ε = δ we have
|x| < δ ⇒ |f(x) – 7| < ε
∴ \(\lim _{x \rightarrow 0}\) (7) = 7

Question 6.
\(\lim _{x \rightarrow 1} \frac{(x-1)^3}{(x-1)^3}\)
Solution:
We guess the limit is 1.
Let us confirm using ε – δ technique.
Let ε > 0, f(x) = \(\frac{(x-1)^3}{(x-1)^3}\)
We will choose a δ > 0 such that
|x – 1| < δ ⇒ |f(x) – 1)| < ε
Now |f(x) – 1| < ε
if 1 – ε < f(x) < 1 + ε
∴ We will choose a δ > 0 such that
x ∈ (1 – δ, 1 + δ) – { 1 }
⇒ f(x) ∈ ( 1 – ε, 1 + ε)
As f(x) = for x ≠ 1
We have f(x) ∈ (1 – ε. 1 + ε) for all x ∈ (1 – δ, 1 + δ) – [1]
∴ We can choose δ = ε
for given ε > 0, there exists δ = ε
s.t. |x – 1| < δ ⇒ |f(x) – 1| < ε
∴ \(\lim _{x \rightarrow 1}\) f(x) = 1

Question 7.
\(\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}\)
Solution:
If we take x very close to 3 (≠ 3)
we have \(\frac{x^3-9}{x-3}\)
= \(\frac{(x-3)\left(x^2+3 x+3^2\right)}{2}\) → 27
Let ε > 0 and x ≠ 3
Now |\(\frac{x^3-9}{x-3}\) – 27| = |x² + 3x +9 – 27|
=|x² – 9 + 3(x – 3)| ≤ |x² – 9| + 3|x – 3|
= |x – 3| [|x + 3| + 3] ≤ |x – 3| [|x + 6| < |x – 3| [|x – 3 + 9|]]
If |x – 3| < δ and δ < 1 then |x – 3| [x – 3 + 9| < δ {1 + 9} = 10 δ
Let δ = min {1, \(\frac{\varepsilon}{10}\)}
∴ For given ε > 0 we have a δ = min {1, \(\frac{\varepsilon}{10}\)} >0 such that
|x – 3| < δ ⇒ |\(\frac{x^3-9}{x-3}\) – 27|
∴ \(\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}\) = 27

Question 8.
\(\lim _{x \rightarrow 1} \frac{3 x+2}{2 x+3}\)
Solution:
we observe that as x → 1, \(\frac{x+2}{2 x+3}\) → 1
To establish this let ε > 0,
we seek a δ > 0,

Question 9.
\(\lim _{x \rightarrow 0}|x|\)
Solution:
We see that when x → 0,|x| → 0
Let us establish this using ε – δ technique.
Let ε > 0 we seek a δ > 0 depending on
ε s.t.|x – 0| < ε ⇒ ||x| – 0| < ε
Now ||xl – 0| = ||x|| = |x| < δ
By choosing ε = δ we have |x| < ε ⇒ ||x| – 0| < ε
∴ \(\lim _{x \rightarrow 0}|x|\) = 0

Question 10.
\(\lim _{x \rightarrow 2}(|x|+3)\)
Solution:
We see that as x → 2, |x| + 3 → 5
Let ε > 0 we were searching for a, δ > 0
such that |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
Now ||x|| + 3 – 5| = ||x| – 2| < |x – 2| < δ
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ \(\lim _{x \rightarrow 2}(|x|+3)\) = 5

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Ex 12(a) Textbook Exercise Questions and Answers.

CHSE Odisha Class 11 Math Solutions Chapter 12 Conic Sections Exercise 12(a)

Question 1.
Fill in the blanks by choosing the correct answer from the given alternatives :
(a) The center of the circle x² + y² + 2xy – 6y + 1 = 0 is _____________. [(2, -6), (-2, 6), (-1, 3), (1, -3)]
Solution:
(-1, 3)

(b) The equation 2x² – ky² – 6x + 4y – 1 = 0 represents a circle if k = ____________. [2, -2, 0, 1]
Solution:
-2

(c) The point (-3, 4) lies ______________ the circle x² + y² = 16 [outside, inside, on]
Solution:
Outside

(d) The line y = x + k touches the circle x² + y² = 16 if k = _______________. [±2√2, ±4√2, ±8√2, ±16√2]
Solution:
±4√2

(e) The radius of the circle x² + y² – 2x + 4y + 1 = 0 is _______________. [1, 2, 4, √19]
Solution:
2

Question 2.
State (with reasons), which of the following is true or false :
(a) Every second-degree equation in x and y represents a circle.
Solution:
Every 2nd-degree equation in x and y represents a circle if the coefficients of x and y are equal and the equation does not contain xy term (False)

(b) The circle (x – 1)² + (y – 1)² = 1 passes through origin.
Solution:
(0 – 1)² + (0 – 1)² = 1 + 1 = 2 ≠ 1.
So the circle does not pass through the origin. (False)

(c) The line y = 0 is a tangent to the circle (x + 1)² + (y – 2)²  = 1.
Solution:
The line y = 0 is a tangent to the circle centre at (-1, 2) and the radius is 1. (True)
∴ The distance of the centre from the line y = 0 is 1 which is equal to its radius.

(d) The radical axis of two circles always passes through the centre of one of the circles,
Solution:
As radical axis is the common chord of the circles, which should not pass through the centre of one of the circles. (False)

(e) The circle x² + (y – 3)² = 4 and (x – 4)² + y² = 9 touch each other.
Solution:
The distance between the centres is \(\sqrt{(0-4)^2+(3-0)^2}\) = 5 which is equal to the sum of the radii. (True)

Question 3.
Find the equation of circles determined by the following conditions.
(a) The centre at (1, 4) and passing through (-2, 1).
Solution:

(b) The centre at (-2, 3) and passing through origin.
Solution:
Centre at (-2, 3) and circle passes through origin.
∴ Radius of the circle = \(\sqrt{(-2)^2+3^2}=\sqrt{13}\)
∴ Equation of the circle is (x – h)² + (y – k)² = a²
or, (x + 2)² + (y – 3)² = 13

(c) The centre at (3, 2) and a circle is tangent to x – axis.
Solution:

(d) The centre at (-1, 4) and circle is tangent to y – axis.
Solution:

(e) The ends of diameter are (-5, 3) and (7, 5).
Solution:
The endpoints of the diameter of the circle are (-5, 3) and (7, 5).
∴ Equ. of the circle is
(x – h)² + (y – k)² = a²
(x- x₁)(x – x₂) + (y – y₁)(y – y₂) = 0
or, (x + 5)(x – 7) + (y – 3)(y – 5) = 0
or, x² – 7x + 5x – 35 + y² – 5y – 3y + 15 = 0
or, x² + y² – 2x – 8y – 20 = 0

(f) The radius is 5 and circle is tangent to both axes.
Solution:
As the circle is tangent to both axes, we have its centre at (5, 5).
∴ Equation of the circle is
or, (x ± 5)² + (y ± 5)² = 25

(g) The centre is on the x-axis and the circle passes through the origin and the point (4, 2).
Solution:

∴ \(\sqrt{(4-a)^2+4}\) = a
or, (4 – a)² + 4 = a²
or, 16 + a² – 8a + 4 = a²
or, 8a = 20 or, a = \(\frac{20}{8}=\frac{5}{2}\)
∴ Equation of the circle is
(x – h)² + (y – k)² = a²
or, (x – \(\frac{5}{2}\))² + (y – 0)² = (\(\frac{5}{2}\))²
or, x² + \(\frac{25}{4}\) – 5x + y² = \(\frac{25}{4}\)
or, x² + y² – 5x = 0

(h) The centre is on the line 8x + 5y = 0 and the circle passes through the points (2, 1) and (3, 5).
Solution:
Let the equation of the circle be x² + 2gx + y² + 2fy + c = 0
∴ Its centre at (- g, -f). As the centre lies on the line 8x + 5y = 0
We have -8g – 5f = 0      …..(1)
Again, as the circle passes through points (2, 1) and (3, 5)
We have
4 + 4g + 1 + 2f + c = 0  …..(2)
and 9 + 6g + 25 + 10f + c = 0   …..(3)
Now from (1), we have g = \(\frac{-5 f}{8}\)
From equation (2), 4g + 2f + c + 5 = 0
or, 4 \(\frac{-5 f}{8}\) + 2f + c + 5 = 0
or, -5f + 4f + 2c + 10 = 0
or, f = 2c + 10    …..(4)
(2) 6g + 10f + c + 34 = 0
or, 6\(\frac{-5 f}{8}\) + 10f + c + 34 = 0
or, -15f + 40f + 4c + 136 = 0
or, 25f = -4c – 136
or, f = \(\frac{-4 c-136}{25}\)
∴ 2c + 10 = \(\frac{-4 c-136}{25}\)
or, 25 (c + 5) = -2c – 68
or, 25c + 2c = -68 – 125
or, 27c = -193 or, c = \(\frac{-193}{27}\)
∴ f = 2C + 10 = 2(\(\frac{-193}{27}\)) + 10
= \(\frac{-386+270}{27}=\frac{-116}{27}\)
∴ g = \(\frac{-5 f}{8}=\frac{-5}{8} \times\left(\frac{-116}{27}\right)=\frac{145}{54}\)
Eqn. of the circle is x² + y² + 2 × \(\frac{145}{54}\) x + 2 \(\frac{-116}{27}\) y + \(\frac{-193}{27}\) = 0
or, 27x² + 27y² + 145x – 232y – 193 = 0

(i) The centre is on the line 2x + y – 3 = 0 and the circle passes through the points (5, 1) and (2, -3).
Solution:
Let the eqn. of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through (5, 1) and (2, -3),
we have 25 + 1 + 10g + 2f + c = 0 …(1)
and 4 + 9 + 4g – 6f + c = 0    …..(2)
Again as the centre lies on the line 2x + y – 3 = 0,
we have- 2g – f – 3 = 0 or, f= -2g – 3
∴ From equation (1)
10g + 2 (-2g – 3) + c + 26 = 0
or, 10g – 4g – 6 = -c – 26
or, 6g = -c – 20
or, g = \(\frac{-c-20}{6}\)
∴ From equation (2)
4g – 6 (-2g – 3) + c + 13 = 0
or, 4g + 12g + 18 + c + 13 = 0
or, 16g = -c – 31
or, g = \(\frac{-c-31}{16}\)
∴ \(\frac{-c-20}{6}=\frac{-c-31}{16}\)
or, -8c – 160 = -3c – 93
or, 5c = -160 + 93 = -67
or, c = –\(\frac{67}{5}\)
∴ g = \(\frac{-c-20}{6}=\frac{\frac{67}{5}-20}{6}=\frac{67-100}{5 \times 6}\)
= \(\frac{-33}{5 \times 6}=\frac{-11}{10}\)
∴ f = -2g – 3 = (-2)\(\left(\frac{-11}{10}\right)\)
= \(\frac{11-15}{5}=\frac{-4}{5}\)
∴ Eqn. of the circle is x² + y² + 2 (\(\frac{-11}{10}\))x + 2(\(\frac{-4}{5}\))y – \(\frac{67}{5}\) = 0
or, 5x² + 5y² – 11x – 8y – 67 = 0

(j) The circle is tangent to the line x + 2y – 9 = 0 at (5, 2) and also tangent to the line 2x – 3y – 7 = 0 at (2, -1).
Solution:

(k) The circle touches the axis of x at (3, 0) and also touches the line 3y – 4x = 12.
Solution:
Let the centre be at (3, k)
Radius = k

or, 3k – 24 = ±5k or, 2k = -24
or, k = -12
Also k = 3
∴ Equation of the Circle is
(x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y2 – 6x – 2 (-12)y = 0
or, x² + y² – 6x + 24y + 9 = 0
and x² + y² – 6x – 6y + 9 = 0

(l) Circle is tangent to x – axis and passes through (1, -2) and (3, -4).
Solution:
Let the centre be at (h, k).
So the radius is k.

∴ Equation of the circle is (x – h)² + (y – k)² = k²
or, (x + 5)² + (y + 10)² = 100 and (x – 3)² + (y + 2)² = 4

(m) Circle passes through origin and cuts of intercepts a and b from the axes.
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(n) Circle touches the axis of x at a distance of 3 from the origin and intercepts a distance of 6 on the y-axis.
Solution:
Let the centre be at (3, k).
So the radius is k.
∴ Equation of the circle is (x – 3)² + (y – k)² = k²
or, x² + 9 – 6x + y² + k² – 2ky = k²
or, x² + y² – 6x – 2xy + 9 = 0

∴ |y₂ – y₁| = 2\(\sqrt{k^2-9}\) = 6
or, \(\sqrt{k^2-9}\) = 3
or, k² = 18, or, k = ±3√2
∴ Equation of the circle is x² – y² – 6x ± 6y√2 + 9 = 0

Question 4.
Find the centre and radius of the following circles:
(a) x² + y² + 6xy – 4y – 12 = 0
Solution:
x² + y² + 6xy – 4y – 12 = 0
∴ 2g = 6, 2f = – 4, c = -12
∴ 8 = 3, f = -2
Centre of (-g, -f) = (-3, 2) and radius = \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{9+4+12}\) = 5

(b) ax² + ay² + 2gx + 2fy + k = 0
Solution:
ax² + ay² + 2gx + 2fy + k = 0
or, x² + y² + \(\frac{2 g}{a}\)x + \(\frac{2 f}{a}\)y + \(\frac{k}{a}\) = 0
∴ Centre of \(\left(\frac{-g}{a}, \frac{-f}{a}\right)\)
and radius = \(\sqrt{\frac{g^2}{a^2}+\frac{f^2}{a^2}-\frac{k}{a}}=\sqrt{\frac{g^2+f^2-a k}{a}}\)

(c) 4x² + 4y² – 4x + 12y – 15 = 0
Solution:
4x² + 4y² – 4x + 12y – 15 = 0
or, x² + y² – 4 + 3y  – \(\frac{15}{4}\) = 0
∴ 2g = -1, 2f = 3, c = \(\frac{15}{4}\)
∴ g = – \(\frac{1}{2}\), f = \(\frac{3}{2}\)
∴ Centre at (-g, -f) = (\(\frac{1}{2}\), \(\frac{-3}{2}\)) and radius \(\sqrt{g^2+f^2-c}\)
= \(\sqrt{\frac{1}{4}+\frac{9}{4}+\frac{15}{4}}=\frac{5}{2}\)

(d) a(x² + y²) – bx – cy = 0
Solution:
a(x² + y²) – bx – cy = 0
or, x² + y² – \(\frac{b x}{a}\) – \(\frac{c y}{a}\) = 0

Question 5.
Obtain the equation of circles passing through the following points and determine the coordinates of the centre and radius of the circle in each case:
(a) the points (3, 4) (4, -3) and (-3, 4).
Solution:
Let the centre be at (h, k)

(b) the points (2, 3), (6, 1) and (4, -6).
Solution:
Let the centre be at (h, k).

we have \(|\overline{\mathrm{PC}}|=|\overline{\mathrm{QC}}|=|\overline{\mathrm{RC}}|\)
∴ (h – 4)² + (k + 6)² = (h – 6)² + (k – 1)² and (h – 4)² + (k + 6)² = (h – 2)² + (k – 3)²
∴ h² + 16 – 8h + k² + 36 + 12k
= h² + 36 – 12h + k² + 1 – 2k
and h² + 16 – 8h + k² + 36 + 12k
= h² + 4 – 4h + k² + 9 – 6k
or, 14k = -4h – 15 and 18k = 4h – 39
or, k = \(\frac{-4 h-15}{14}\) and k = \(\frac{4 h-39}{18}\)

(c) the points (a, 0), (-a, 0) and (0, b).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0. As it passes through the points (a, 0), (-a, 0) and (0, b). We have
a² – 2ga + c = 0   …..(1)
a² + 2ga + c = 0   …..(2)

(d) the points (-3, 1), (5, -3) and (-3, 4).
Solution:
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points. we have (-3, 1), (5, -3) and (3, 4).
We have 9 + 1 – 6g + 2f + c = 0    …..(1)
25 + 9 + 10g – 6df + c = 0      …(2)
9 + 16 – 6g + 8f + c = 0     …..(3)

Question 6.
Find the equation of the circles circumscribing the triangles formed by the lines given below :
(a) the lines x = 0, y = x, 2x + 3y = 10
Solution:

∴ The coordinates. C are (0, 0) of
Lastly, solving \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{AC}}\)
we have y = x, 2x + 3y = 1 0
we have 5x = 10
or, x = 2 and y = 2.
∴ The coordinates of A are (2, 2).
∴ The circle passes through the points (2, 2), (0, \(\frac{10}{3}\)) and (0, 0)
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0
As it passes through the points A, B, C we have c = 0, 4 + 4 + 4g + 4f = 0,
\(\frac{100}{9}\) + 2. f. \(\frac{10}{3}\) + 0 = 0
∴ f = \(\frac{-100}{9} / \frac{20}{3}=\frac{-5}{3}\)
and g = \(\frac{-4 f-8}{4}=\frac{-4\left(\frac{-5}{3}\right)-8}{4}\)
= \(\frac{20-24}{3 \times 4}=\frac{-1}{3}\)
∴ Equation of the circle is  x² + y² + 2(\(\frac{-1}{3}\))x + 2 \(\frac{-5}{3}\)y + 0 = 0
or, 3(x² +  y²) – 2x – 10y = 0

(b) The lines x = 0, 4x + 5y = 35, 4y = 3x + 25
Solution:



or, 4x² + 4y² – 24x – 53y + 175 = 0

(c) The lines x = 0, y = 0, 3x + 4y – 12 = 0
Solution:
The coordinates of A, B and C are (4, 0), (0, 3) and (0, 0).
Let the equation of the circle be x² + y² + 2gx + 2fy + c = 0

(d) The lines y = x, y =2 and y = 3x + 2
Solution:

(e) the lines x + y = 6, 2x + y = 4 and x + 2y = 5
Solution:

Question 7.
Find the coordinates of the points where the circle x² + y² – 7x – 8y + 12 = 0 meets the coordinates axes and hence find the intercepts on the axes. [Hint: If a circle intersects a line at points A and B, then the length AB is its intercepts on line L]
Solution:
x² + y² – 7x – 8y + 12 = 0
Putting x = 0, we have y² – 8y + 12 = 0 or, (y – 6) (y – 2) = 0, or, y = 6, 2.
∴ The circle meets the Y-axis at (0, 6) and (0, 2) and its Y-intercept is 6 – 2 = 4.
Again putting y = 0,
we have x² – 7x + 12 = 0
or, (x – 4)(x – 3) = 0 or, x = 4, x = 3.
∴ The circle meets the X-axis at (4, 0) and (3, 0) and its x-intercept is 4 – 3 = 1.

Question 8.
Find the equation of the circle passing through the point (1, -2) and having its centre at the point of intersection of lines 2x – y + 3 = 0 and x + 2y – 1 =0
Solution:

Question 9.
Find the equation of the circle whose ends of a diameter are the points of intersections of the lines and x + y – 1 = 0, 4x + 3y + 1 = 0 and 4x +y + 3 = 0, x – 2y +3 = 0.
Solution:
Solving x + y – 1 = 0, 4x + y + 3 = 0

∴ The endpoints of the diameter are (-4, 5) and (-1, 1).
∴ Equation of the circle is
(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
or, (x + 4) (x + 1) + (y – 5) (y – 1) = 0
or, x² + x + 4x + 4 + y² – y – 5y + 5 = 0
or, x² + y² + 5x – 6y + 9 = 0.

Question 10.
Find the equation of the circle inscribed inside the triangle formed by the line \(\frac{x}{4}+\frac{y}{3}\) = 1 and the coordinate axes.
Solution:
The circle is inscribed in the triangle formed by x = 0, y = 0 and \(\frac{x}{4}+\frac{y}{3}\) = 1
∴ If (h, k) is the centre and r is the radius of the circle then h = k = r.
The perpendicular distance of the centre (h, h) from the line 3x + 4y = 12 is the radius.
⇒ \(\left|\frac{3 h+4 h-12}{5}\right|\) = h
⇒ 7h – 12 = ±5h
⇒ 2h = 12 or 2h = 12
⇒ h = 6 or h = 1
But h can not be 6 thus the circle has equation (x – 1)² + (y – 1)² = 1
⇒ x² + y² – 2x – 2y + 1 =0

Question 11.
(a) Find the equation of the circle with its centre at (3, 2) and which touches to the line x + 2y – 4 = 0.
Solution:

(b) The line 3x + 4y + 30 = 0 is a tangent to the circle whose centre is at (\(-\frac{12}{5},-\frac{16}{5}\)). Find the equation of the circle.
Solution:

(c) Prove that the points (9, 7), and (11, 3) lie on a circle with centre at origin. Find the equation of the circle.
Solution:

(d) Find the equation of the circle which touches the line x = 0, x = a and 3x + 4y + 5a = 0.
Solution:

(e) If a circle touches the co-ordinate axes and also touches the straight line \(\frac{x}{a}+\frac{y}{b}\) = 1 and has its centre in the 1st quadrant, And its equation.
Solution:
Let the centre be at (k, k) and the radius is k.

Question 12.
ABCD is a square of side ‘a’ If AB and AD are taken as co-ordinate axes, prove that the equation of the circle circumscribing the square is
x² + y² = a(x + y)
Solution:

or, x² + \(\frac{a^2}{4}\) – ax + y² + \(\frac{a^2}{4}\) – ay = \(\frac{a^2}{2}\)
or, x² + y² – ax – ay = 0
or, x² + y² = a(x + y)

Question 13.
(a) Find the equation of the tangent and normal to the circle x² + y² = 25 at the point (3, -4).
Solution:
Equation of the tangent to the circle x² + y² = 25 at the point (3, -4) is
xx₁ + yy₁ = a²
3x – 4y = 25
Equation of the normal is x₁y = xy₁
or, 3y = -4x or, 4x + 3y = 0

(b) Find the equation of the tangent and normal, to the circle, x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3).
Solution:
Equation of the tangent of the circle x² + y² – 3x + 4y – 31 = 0 at the point (-2, 3) is
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, -2x + 3y – \(\frac{3}{2}\) (x – 2) + 2(y + 3) – 31 = 0
or, – 4x + 6y – 3x + 6 + 4y + 12 – 62 = 0
or, -7x + 10y – 44 = 0
or, 7x – 10y + 44 = 0
Equation of the normal is x(f + y₁) – y(g + x₁) fx₁ + gy₁ = 0
or, x(2 + 3) – y(\(-\frac{3}{2}\) – 2) – 2(-2) – \(\frac{3}{2}\) × 3 = 0
or, 5x + \(\frac{7y}{2}\) + 4 – \(\frac{9}{2}\) = 0
or, 10x + 7y – 1 = 0

(c) Find the equation of the tangents to the circle x² + y² + 4x – 6y – 16 = 0 at the point where it meets the y – axis.
Solution:
Putting x = 0 in the circle equation, we have
y² – 6y – 16 = 0
or, y² – 8y + 2y – 16 = 0
or, y(y – 8) + 2(y – 8) = 0
or, (y – 8)(y + 2) = 0
y = 8 or, -2
The circle meets y – axis at (0, 8) and (0, -2).
Eqn. of the tangents are
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, 0 + 8y + 2 (x + 0) – 3(y + 8) – 16 = 0
or, 8y + 2x – 3y – 24 – 16 = 0
or, 2x + 5y = 40 and
x × 0 – 2y + 2 (x + 0) – 3 (y – 2) – 16 = 0
or, -2y + 2x – 3y + 6 – 16 = 0
or, 2x – 5y – 10 = 0

(d) Find the condition under which the tangents at (x₁, y₁) and (x₂, y₂) to the circle x² + y² + 2gx + 2fy + c = 0 are perpendicular.
Solution:
Equation of tangent to the circle
x² + y² + 2gx + 2fy + c = 0 at the point (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
or, (g + x₁)x + y(f + y₁) + gx₁ + fy₁ + c = 0
Again equation of the tangent to the circle at (x₂, y₂) is
x(g + x₂) + y(f + y₂) + gx₂ + fy₂ + c = 0
As the tangent (1) and (2) are perpendicular, we have the product of their slopes is -1.
∴ \(\frac{g+x_1}{f+y_1} \times \frac{g+x_2}{f+y_1}\) = -1
or, (g + x₁)(g + x₂) = -(f + y₁)(f + y₂)
or, (g + x₁)(g + x₂) + (f + y₁)(f + y₂) = 0

(e) Calculate the radii and distance between the centres of the circles, whose equations are, x² + y² – 16x – 10y + 8 = 0; x² + y² + 6x – 4y – 36 = 0. Hence or otherwise prove that the tangents drawn to the circles at their points of intersection are perpendicular.
Solution:
x² + y² – 16x – 10y + 8 = 0;
x² + y² + 6x – 4y – 36 = 0.
g₁ = -8, f₁ = -5, c₁ = 8,
g₂ = 3, f₂ = -2, c₂ = -36
The centres are (-g₁, -f₁) and (-g₂, -f₂)

Question 14.
(a) Find the equation of the tangents to the circle x² + y² = 9 perpendiculars to the line x – y – 1 = 0
Solution:

(b) Find the equation of the tangent to the circle x² + y² – 2x – 4y = 40, parallel to the line 3x – 4y = 1.
Solution:


(c) Show that the line x – 7y + 5 = 0 a tangent to the circle x² + y² – 5x + 5y = 0. Find the point of contact. Find also the equation of tangent parallel to the given line.

Solution:
we have the line is x – 7y + 5 = 0
or, y = \(\frac{x+5}{7}\)
Now putting the value of y in the circle, we have x² + y² – 5x + 5y = 0
or, x² + (\(\frac{x+5}{7}\))² – 5x + 5 \(\frac{x+5}{7}\) = 0
or, 49x² + x² + 25 + 10x – 245x + 35x + 175 = 0
or, 50x² – 200x + 200 = 0
or, x² – 4x + 4 = 0
∴ a = 1, b = -4, c = 4
∴ b² – 4ac = (-4)² – 4 × 1 × 4
= 16 – 16 = 0
∴ The line x – 7y + 5 = 0

x – 7y – 45 and x – 7y + 5 = 0

(d) Prove that the line ax + by + c = 0 will be the tangent to the circle x² + y² = r² if r²(a² + b²) = c².
Solution:
We know that a line is a tangent to the circle if the distance of the line from the centre is equal to the radius.
Now the circle is x² + y² = r²
⇒ Centre is at (0, 0) and radius r. The distance of (0, 0) from ax + by + c = 0 is

(e) Prove that the line 2x + y = 1 tangent to the circle x² + y² + 6x – 4y + 8 = 0.
Solution:

(f) If the line 4y – 3x = k is a tangent to the circle x² + y² + 10x – 6y + 9 = 0 find ‘k’. Also, find the coordinates of the point of contact.
Solution:
Center of the circle is (-5, 3) and the radius is \(\sqrt{25+9-9}\) = 5
Distance of the centre from the line 4y – 3x – k = 0

Question 15.
(a) Find the length of the tangent, drawn to the circle x² + y² + 10x – 6y + 8 = 0 from the centre of the circle x² + y² + 4x = 0.
Solution:
Center of the circle x² + y² + 4x = 0 is (2, 0)
∴ Length of the tangent drawn from the point (2, 0) to the circle x² + y² + 10x – 6y + 8 = 0
\(\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 g \mathrm{~g}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{4+0+10 \times 2+0+8}=\sqrt{32}=4 \sqrt{2}\)

(b) Find the length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0
Solution:
Length of the tangent drawn from the point (2, -1) to the circle x² + y² + 6x + 10y + 18 = 0 is
\(\sqrt{\mathrm{x}_1{ }^2+\mathrm{y}_1{ }^2+2 \mathrm{gx}_1+2 \mathrm{fy}_1+\mathrm{c}}\)
= \(\sqrt{4+1+(-6) \times 2+10(-1)+18}\)
= \(\sqrt{5-12-10+18}\) = 1

(c) Find the length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15.
Solution:
Length of the tangent drawn from the point (4, 7) to the circle x² + y² = 15 is \(\sqrt{16+49-15}\) = √50 = 5√2

Question 16.
(a) Prove that the circle given by the equations x² + y² + 2x – 8y + 8 = 0 and x² + y² + 10x – 2y + 22 = 0 touches each other externally. Find also the point of contact
Solution:
x² + y² + 2x – 8y + 8 = 0
g₁ = 1, f₁ = -4, c₁ = 8
Hence centre = c₁(-g₁, -f₁) = c₁(-1, 4)
Radius = r₁ = \(\sqrt{1+16-8}\) = 3
Again x² + y² + 10x – 2y + 22 = 0
g₂ = 5, f₂ = -1, c₂ = 22
Centre c₂(-g₂, -f₂) = c₂(-5, 1)
Radius r₂ = \(\sqrt{25+1-22}\) = 2
Now

(b) Prove that the circle is given by the equations x² + y² = 4 and x² + y² + 6x + 8y – 24 = 0, touch each other and find the equation of the common tangent.
Solution:
x² + y² = 4,
x² + y² + 6x + 8y – 24 = 0
Their centres are (0, 0) and (-3, -4) and radii are 2 and \(\sqrt{9+16+24}\) = 7
∴ Distance between the centres is \(\sqrt{(-3)^2+(-4)^2}\) = 5, which is equal to the difference between the radii.
∴ The circles touch each other internally.
∴ Equation of the common tangent is S₁ – S₂ = 0
or, (x² + y² + 6x + 8y – 24) – (x² + y² – 4) = 0
or, 6x + 8y – 20 = 0
or, 3x + 4y = 10

(c) Prove that the two circle x² + y² + 2by + c² = 0 and x² + y² + 2ax + c² = 0,  will touch each other \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}\).
Solution:
x² + y² + 2by + c² = 0,
x² + y² + 2ax + c² = 0
g₁ = 0, f₁ = b, c₁ = c².
g₂ = a, f₂ = 0, c₂ = c².
The centres of the circle are (0, -b) and (-a, 0) and radii are \(\sqrt{b^2-c^2}\) and \(\sqrt{a^2-c^2}\). As the circles touch each other, we have the distance between the centres is equal to the sum of the radii.

(d) Prove that the circles given by x² + y² + 2ax + 2by + c = 0, and x² + y² + 2bx + 2ay + 2c = 0, touch each other, if (a + b) = 2c.
Solution:
x² + y² + 2ax + 2by + c = 0
x² + y² + 2bx + 2ay + 2c = 0,
The centre of the circle is (-a, -b) and (-b, -a). the radii of the circle are \(\sqrt{a^2+b^2-c}\) and \(\sqrt{b^2+a^2-c}\). As the circles touch each other we have, the distance between the centres is equal to the sum of the radii.

Question 17.
Find the equation of the circle through the point of intersection of circles x² + y² – 6x = 0 and x² + y² + 4y – 1 = 0 and the point (-1, 1).
Solution:
Let the equation of the circle be (x² + y² – 6x) + λ(x² + y² + 4y – 1) = 0
As it passes through the point (-1, 1),
we have (1 + 1 + 6) + λ(1 + 1 + 4 – 1) = 0
or, 8 + 5λ = 0 or, λ = \(\frac{-8}{5}\)
∴ Equation of the circle is (x² + y² – 6x) – \(\frac{8}{5}\) (x² + y² + 4y – 1) = 0
or, 5x² + 5y² – 30x – 8x² – 8y² – 32y + 8 = 0
or, 3x² + 3y² + 30x + 32y – 8 = 0

Question 18.
Find the equation of the circle passing through the intersection of the circles, x² + y² – 2ax = 0 and x² + y² – 2by = 0 and having the centre of the line \(\frac{x}{a}-\frac{y}{b}\) = 2
Solution:

Question 19.
Find the radical axis of the circles x² + y² – 6x – 8y – 3 = 0 and 2x² + 2y² + 4x – 8y = 0
Solution:
x² + y² – 6x – 8y – 3 = 0
2x² + 2y² + 4x – 8y = 0
x² + y² – 6x – 8y – 3 = 0
x² + y² + 2x – 4y = 0
∴ The equation of the radical axis is S₁ – S₂ = 0
or, (x² – y²– 6x – 8y – 3) – (x² + y² + 2x – 4y) = 0
or, -6x – 8y- 3 – 2x + 4y = 0
or, -8x – 4y – 3 = 0
or, 8x + 4y + 3 = 0

Question 20.
Find the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0 Prove that the radical axis is perpendicular to the line joining the centres of the two circles.
Solution:
Equation of the radical axes of the circle x² + y² – 6x + 8y – 12 = 0  and x² + y² + 6x – 8y + 12 = 0
(x² + y² – 6x + 8y – 12) – (x² + y² + 6x – 8y + 12) = 0
or, -12x + 16y – 24 = 0
or, 3x – 4y + 6 = 0
Again, slope of the radical axis is \(\frac{3}{4}\) = m₁ (say)
Centres of the circles are (3, -4) and (-3, 4).
Slope of the line joining the centres is \(\frac{4+4}{-3-3}=\frac{8}{-6}=-\frac{4}{3}\) = m₂ (say)
m₁. m₂ = \(\frac{3}{4}\left(-\frac{4}{3}\right)\) = -1
∴ The radical axis is perpendicular to the line joining centres of the circles. (Proved)

Question 21.
If the centre of one circle lies on or inside another, prove that the circles cannot be orthogonal.
Solution:
The orthogonality condition for two circles.
x² + y² + 2g₁x + 2f₁y + C₁ = 0   …..(1)
and x² + y² + 2g₂x + 2f₂y + C₂ = 0    …..(2)
is 2(g₁g₂ + f₁f₂) – C₁ – C₂ = 0
Let us consider two circles
Case-1. Let the centre of (2) which is C (-g₂, -f₂) lies on the circle (1). Hence it satisfies the equation (i)
i.e., g₂² + f₂² – 2g₁g₂ – 2f₁f₂ + C₂ = 0
⇒ 2g₁g₂ + 2f₁f₂ – C₁ – C₂ = g₂² + f₂² – C₂
Its right-hand side is the square of the radius of 2nd circle which can not be equal
to zero i.e., 2(g₁g₂ + f₁f₂) – C₁ – C₂ ≠ 0
Hence circles are not orthogonal.
Case-2. Let the centre of (2) which is (-g2, -f2) lies inside the circle (1).
Distance between their centres < radius of the first circle.
i,e. \(\sqrt{\left(g_1-g_2\right)^2+\left(f_1-f_2\right)^2}<\sqrt{g_1^2+f_1^2-C_1}\)
⇒ g₁² – 2g₁g₂ + f₁² + f₂² + 2f₁f₂ < g₁² + f₁² – C₁
⇒ 2g₁g₂ – 2f₁f₂ – C₁ – C₂ > g₂² + f₂² – C₂
= square of the radius of 2^nd circle. Hence greater than 0.
⇒ 2(g₁g₂ + f₁f₂) – C₁ – C₂ > 0
So two circles are not orthogonal. By case -1 and case -2 we conclude that if the centres of one circle lie on or inside another, then circles cannot be orthogonal.

Question 22.
If a circle S intersects circles S₁ and S₂ orthogonally. Prove that the centre of S lies on the radical axis of S₁ and S₂. [Hints: Take the line of centres of S₁ and S₂ as x – axis and the radical axis as y – axis. Use conditions for the orthogonal intersection of S, S₁ and S, S₂ simultaneously and prove that S is centred on the y – axis.]
Solution:
Let the equation of the circle S, S₁ and S₂ are
x² + y² + 2gx + 2fy + C = 0      …(1)
x² + y² + 2g₁x + 2f₁y + C = 0      …(2)
and x² + y² + 2g₂x + 2f₂y + C = 0      …(3)
According to the question, the circle S intersects circles S₁ and S₂ orthogonally.
Hence 2 (g₁g + f₁f) – C₁ – C = 0 …(4)
and 2 (g₂g + f₂f) – C₂ – C = 0  ….(5)
Subtracting (4) from (3) we get
2g(g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0 …(6)
Now radical axis of circles S₁ and S₂ is S₁ – S₂ = 0
i, e. 2x (g₁ – g₂) + 2y (f1 – f2)+ C₁ – C₂ = 0 ….(7)
The centre of the circle S is (-g, -f).
If it lies in the radical axis then equation (7) will be satisfied by the centre.
i,.e, 2g (g₁ – g₂) + 2f (f₁ – f₂) – C₁ + C₂ = 0
which is nothing but equation (5). Hence centres of S lie on the radical axis of S₁ and S₂.

Question 23.
R is the radical centre of circles S₁, S₂ and S₃. Prove that if R is on/inside/outside one of the circles then it is similarly situated with respect to the other two.
Solution:
Given R is the radical centre of S₁, S₂ and S₃
The radical centre is the intersection point of three radical axes whose equations are
S₁ – S₂ = 0
S₂ – S₃ = 0    …..(1)
S₃ – S₁ = 0
Let S₁ : x² +y² + 2g₁x + 2f₁y + C₁ =0
S₂ : x² + y² + 2g₂x + 2f₂y + C₂ =0
S₃ : x² + y² + 2g₃x + 2f₃y + C₃ =0
Now equations of radical axes by set of equation (1) are
2x(g₁ – g₂) + 2y(f₁ – f₂) + C₁ – C₂ =0 …(2)
2x(g₂ – g₃) + 2y(f₂ – f₃) + C₂ – C₃ =0 …(3)
and 2x(g₃ – g₁) + 2y(f₃ – f₁) + C₃ – C₁ = 0 …(4)
Let the co-ordinate of R be (x1, y1) the
point R must satisfy (2), (3) and (4).
i.e., 2x₁(g₁ – g₂) + 2y₁(f₁ – f₂) + C₁ – C₂ = 0 …(5)
2x₁(g₂ – g₃) + 2y₁(f₂ – f₃) + C₂ – C₃ =0 …(6)
2x₁(g₃ – g₁) + 2y₁(f₃ – f₁) + C₃ – C₁ =0 …(7)
Subtracting (6) for (5) we get
2x₁(g₁ – g₃) + 2y₁(f₁ – f₃) + C₁ – C₃ =0
⇒ 2g₁x₁ + 2f₁y₁ + C₁ =2g₃x₁ + 2f₃y₁ + C₃
Similarly subtracting (7) from (6) we get
2g₂y₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁
Combining the above two equations we get
2g₂x₁ + 2f₂y₁ + C₂ = 2g₁x₁ + 2f₁y₁ + C₁ = 2g₃x₁ + 2f₃y₁ + C₃
If R x₁ y₁ lies on / inside / outside of S₁ …(8) then x₁² + y₁² + 2g₁x₁ + 2f1y1 + C₂ (= / < / >)0 respectively.
⇒ x₁² + y₁² + 2g₂x₂ + 2f₂y₂ + C₂(=/</>) 0
⇒ x₁² + y₁² + 2g₃x₃ + 2f₃y₃ + C₃(=/</>) 0
respectively by Eqn (8).
This concludes that if R is on /inside/outside. One of the circles then it is similarly situated with respect to the other two.

Question 24.
Determine a circle which cuts orthogonally to each of the circles.
S₁: x² + y² + 4x – 6y + 12 = 0
S₂: x² + y² + 4x + 6y + 12 = 0
S₃: x² + y² – 4x + 6y + 12 = 0
[Hints: The centre of the required circle S must be the radical centre R (why?), which lies outside all the circles. Then show that the radius of S must be the length of the tangent from R to any circle of the system.
Solution:
Let the equation of the required circle is x² + y² + 2gx + 2fy + C = 0    …..(1)
We know if two circles
x² + y² + 2g₁x + 2f₁y + C₂ = 0 and
x² + y² + 2g₂x + 2f₂y + C₂ = 0
are orthogonal then
2(g₁g₂ + f₁f₂) – C₁ – C₂ =0   ….(2)
According to the question circle (1) is orthogonal to the circles
S₁: x² + y² – 4x – 6y + 12 = 0     ….(3)
S₂: x² + y² + 4x + 6y + 12 = 0   ….(4)
S₃: x² + y² – 4x + 6y + 12 = 0     ….(5)
For these circles equation (2) will be
2(-2g – 3f) – C – 12 = 0     ….(6)
2(2g + 3f) – C – 12 = 0     …..(7)
2(-2g + 3f) – C – 12 = 0     …..(8) respectively.
Now subtract eqn. (7) from (6) and (8) from (7) we get
2(- 4g – 6f) = 0
⇒ 2(4g) = 0 ⇒ g = 0 , and f = 0
Using the value of g and f in eq. (6) we get
C = -12
Using g = 0, f= 0 , C = -12 in (1) we get
x² + y² – 12 = 0 is the required equation of the circle.

Question 25.
Prove that no pair of concentric circle can have radical axes.
Solution:
Let the centre of pair of concentric circles is C (h, k) and radii are r₁ and r₂.
So equation of the circles are
S₁: (x – h)² + (y – k)² = r₁²
S₂: (x – h)² + (y – k)² = r₂²
Equation of the radical axis is S₁ – S₂ = 0
⇒ r₁² – r₂² = 0
which is not a straight line as r₁ and r₂ are constants.
Hence it concludes that no pair of concentric circles have a radical axis.

Odisha State Board CHSE Odisha Class 11 Invitation to English 1 Solutions Chapter 3 The Golden Touch Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Chapter 3 The Golden Touch

CHSE Odisha Class 11 English The Golden Touch Text Book Questions and Answers

UNIT – I
Gist with Glossary

Gist:
The legendary king Midas’s madness for gold had no limit. To him, the most precious thing in the world was gold. The love that he had for his little daughter Marygold was no less. It was his earnest wish to see everything such as the golden light of the sun at the evening, a bunch of sweet yellow flowers, and the most beautiful roses in his garden turn into gold. Even the king’s intense love for music in his youth paled into insignificance before the sound of coins, one against another. He always dreamt about gold. He could not resist the temptation of counting his gold pieces.

One morning the sight of an unknown person caught his attention. His astonishment knew no bounds to see him in his treasure room which he himself had locked. He asked Midas, why he was not satisfied, despite being vastly wealthy. He gave vent to his wish that everything he touched would become gold. It would give him supreme happiness. The stranger fulfilled the king’s desire. The following day, at sunrise the latter’s dream came true. He had the Golden Touch at his disposal. Strangely enough, the stranger had vanished.

Glossary:
besides : in addition to (ଏତଦ୍‌ବ୍ୟତୀତ)
dearly : deeply (ଗଭୀର ଭାବରେ)
precious : valuable (ମୂଲ୍ୟବାନ)
treasure-room : here, the room where king Midas had kept bars of gold (ଗନ୍ତାଘର )
brightened : shone (ଉଜ୍ଜ୍ବଳି ଉଠିଲା)
vividly : clearly (ପ୍ରାଞ୍ଜଳ ଭାବରେ)

Think it:
Question 1.
What do you learn about King Midas from the first two paragraphs of the story?
Answer:
The first two paragraphs throw light on King Midas’s vast wealth and his obsession with gold.

Question 2.
What did he wish when saw the golden light of the evening sun?
Answer:
When he saw the golden light of the evening sun, he wished it could change everything into genuine gold.

Question 3.
Why didn’t he like the roses in his garden?
Answer:
He didn’t like the rose in his garden, because they were not made of gold.

Question 4.
How did he spend his time in his ‘treasure room’?
Answer:
He spent his time in his treasure room counting his gold pieces. Besides, he held the bars of gold and praised his gold cups and plates.

Question 5.
How did he come across the stranger?
Answer:
He came across the stranger when his treasure room was bathed in bright sunshine; he found the latter in his locked room.

Question 6.
How did King Midas answer the stranger’s question, “What else do you want ?”
Answer:
When the stranger asked him ‘What else do you want ?’, the king expressed his wish that everything he would touch could be changed into gold. Besides, he was sick of collecting his wealth at a very slow rate.

Question 7.
How did the stranger fulfill his wishes?
Answer:
The stranger fulfilled his ambition by telling him that the following day at sunrise, he would find the Golden Touch at his disposal.

UNIT – II
Gist with Glossary

Gist:
This unit begins with the king’s discovery that his desire for the golden touch had not been fulfilled. His bed had not changed into gold. His sadness did not last long. A sudden sight of the reflection of the golden light of the earliest sunrise on him worked wonders. The sheet on his bed turned into a cloth of gold. The first sunbeam had truly brought the Golden Touch. Driven by excitement, he touched everything – one of the legs of the bed, the curtain at the window, his dress, and spectacles. There was gold everywhere. The loss of spectacles didn’t matter to him much.

The gold was more precious than his pair of spectacles and his daughter Marygold would read to him. The magic touch turned everything into gold, beginning from the brass handle of the door, and the rose trees, the constant source of his pride and joy in the past. At the moment, he went into breakfast that morning, his little beautiful daughter Marygold came in crying bitterly. When he kissed her, she wore a terrible look, with her little face, beautiful hair, and her little body gone. She became a hard golden figure.

Glossary:
turned into gold: transformed into gold (ସୁନା ହୋଇଗଲା)
disappointed : sad (ଦୁଃଖ)
sunbeam : sunlight (ସୂର୍ଯ୍ୟକିରଣ )
put on : wore (ପିନ୍ଧିଲେ)
bitterly: in a painful or unhappy mood (ଦୁଃଖଦ ଅବସ୍ଥାରେ)
scent : fragrance (ସୁଗନ୍ଧ)
comfort : (here) console (ସାନା ଦେବା )
terrible : ferocious (ଭୟଙ୍କର )

Think it out:
Question 1.
When did the king discover that his desire for the golden touch had been fulfilled?
Answer:
The king discovered that his desire for the golden touch had been fulfilled at the sight of his bed sheet transformed into a cloth of gold.

Question 2.
Why was the king not sad when his spectacles turned into gold?
Answer:
The king was not sad when his spectacles turned into gold, because he felt that a pair of spectacles was not as precious as the Golden Touch. Besides, her daughter Marygold could read to him.

Question 3.
What was Marygold’s complaint about the golden rose?
Answer:
Marygold’s complaint about the golden rose was that it had become yellow and hard and lost its fragrance.

Question 4.
How did the king console his daughter?
Answer:
The king consoled his daughter stating that she should not think of it at all because these rose flowers that had turned into gold were invaluable. He lovingly said to her to sit and take her breakfast.

Question 5.
Why couldn’t the king enjoy his breakfast?
Answer:
The king couldn’t enjoy his breakfast which included eggs, fish, bread, butter, and a spoonful of coffee, because they all became gold.

Question 6.
What happened to Marygold when the king kissed her?
Answer:
When the king kissed her, the king saw before him a terrible figure instead of his little daughter. Her sweet face, her beautiful hair, and her little body had all gone. There stood a statue of solid gold.

UNIT-III
Gist with Glossary

Gist:
King Midas sank into despair at the loss of everything he loved including his dear little daughter Marygold. In the meantime, the stranger reappeared and enquired him about his Golden Touch. He narrated his tale of woe to the former. He now realized the futility of the Golden Touch. He was terribly thirsty and pined for a cup of cold water to taste. The stranger kept on asking him what he preferred – the Golden touch or a piece of bread and gold or his own little daughter. The grief-stricken father wanted his daughter to get back. His repentance made the stranger remark that he was wiser than before. The stranger knew human nature dies hard and the king was no different. Midas had now become a virulent hater of gold.

He trembled in fear at the sight of a fly perching on his nose and at once felt the ground turning into a small piece of gold. In order to get rid of the burden of the golden touch, the stranger advised the king to go to the end of his garden, and wash in the water of the river there. This was not all. He should bring some of the same water and sprinkle it over anything, he wished to change back again. The king rose to the occasion without wasting time. To his utter delight, he got his lost daughter back by sprinkling water on her. Marygold was kept in dark about this painful incident. The king started his life afresh in the midst of his loving daughter and his garden full of fragrant roses.

Glossary:
lingered (here) saw for some time (କିଛି ସମୟ ପାଇଁ ଦେଖୁବା)
despair : misery (ଦୁର୍ଦ୍ଦଶା)
passionately : the state of mind caused by passion (ଆଗ୍ର ହାନ୍ତି ତ ହୋଇ)
scrap: piece (ଖଣ୍ଡ)
weight : (here) the burden of greed for gold

Think it out:
Question 1.
How did the king realize that the golden touch was a useless gift for him?
Answer:
The king realized that the golden touch was a useless gift for him because it deprived him of everything that he loved, especially his sweet little daughter Marygold. The king had become a grief-stricken person.

Question 2.
‘You are wiser than you were’ – why did the stranger say so?
Answer:
The stranger said so because he marked that the king was seething with repentance for his obsession with gold. The king wished he had not given one hair of his daughter’s head in exchange for the power to transform the entire earth into gold.

Question 3.
What did the stranger advise the king to do to get rid of his golden touch?
Answer:
In order to get rid of his golden touch, the stranger advised the king to go to the end of his garden, and wash in the water of the river there. This was not all. He should bring some of the same water and sprinkle it over anything he wished to change back again.

Question 4.
How did the king get back his daughter?
Answer:
The king got his daughter back by going straight to the golden figure of his daughter and then sprinkling some water brought from the river on her.

Question 5.
Is the story a tragic or comic one? Give your reasons.
Answer:
The story, The Golden Touch is not a tragic one, because though the grief-stricken king could not look at Marygold, there is no death inevitable. Instead, the story is a comic one. King Midas, the protagonist of the story, in spite of his sadness, makes us laugh at his blind love of gold. How can a father make his loving little daughter a victim of his boundless greed? His reaction at the loss of everything he loved and his belated realization of his mistakes and above all, the stranger’s words contribute to the comic aspect of the story, although there are patches of pathos.

Post-Reading Activities:

A. Arrange the following sentences according to their logical order.
(a) Midas said, ‘I wish everything I touch could be turned into gold’.
(b) ‘The Golden Touch !’ exclaimed the stranger.
(c) Midas said, ‘It would give me perfect happiness’.
(d) The stranger said, ‘Tomorrow at sunrise you will find that you have the Golden Touch’.
(e) King Midas came across a stranger smiling at him.
(f) The stranger asked, ‘What do you want ?’
(g) He guessed that the stranger was no ordinary person.
Answer:
(e) King Midas came across a stranger smiling at him.
(g) He guessed that the stranger was no ordinary person.
(f) The stranger asked, ‘What do you want ?’
(a) Midas said, ‘I wish everything I touch could be turned into gold’.
(b) ‘The Golden Touch !’ exclaimed the stranger.
(c) Midas said, ‘It would give me perfect happiness’.
(d) The stranger said, ‘Tomorrow at sunrise you will find that you have the Golden Touch’.

B. Doing with words.
1. Write the antonyms of the following words :
love —
bright —
perfect —
wise —
please —
usual —
happiness —
common —
beautiful —
careful —
proud —
sincere —
Answer:
love — hate
bright —dull
perfect — imperfect
wise — fool
please — displease
usual — unusual
happiness — sadness
common — uncommon
beautiful — ugly
careful — careless
proud — humble
sincere — insincere

B. Match the expressions in column A with their one-word substitution in column B.

Answer:

3. Write the nouns derived from the following verbs :
collect        ______________
satisfy        ______________
exclaim      ______________
disappoint ______________
reflect        ______________
astonish    ______________
expect       ______________
Answer:
collect        — collection
satisfy        — satisfaction
exclaim      — exclamation
disappoint — disappointment
reflect        — reflection
astonish     — astonishment
expect       — expectation

4. Fill in the blanks with the adjectival forms of the following nouns:
gold        _____________
beauty    _____________
palace     _____________
magic      _____________
spectacle _____________
comfort   _____________
sorrow     _____________
passion    _____________
Answer:
gold        — golden
beauty    — beautiful
palace     — palatial
magic      — magical
spectacle — spectacular
comfort   — comfortable
Sorrow     — sorrowful
passion    — passionate

5. Fill in the blanks with the verbs from which the following nouns have been
speech        _______________
thought      _______________
excitement _______________
collection   ________________
service       _______________
Answer:
speech        — speak
thought      — think
excitement — excite
collection   — collect
service        — serve

CHSE Odisha Class 11 English The Golden Touch Fours Important Questions and Answers

I. Short Answer Type Questions with Answers

1. Read through the extract and answer the questions that follow.
Long ago, there lived a very rich man called Midas. Besides being rich, he was a king, and he had a little daughter called Marygold. King Midas loved gold more than anything else in the world. He liked being a king, chiefly because he loved his golden crown. He loved his daughter dearly too, and the more he loved her, the more gold he wanted for her sake. When King Midas saw the golden light of the sun in the evening, he wished it could turn everything into real gold. When Marygold came to him with a bunch of sweet yellow flowers, he would say, ‘If they were as golden as they look, they would be worth picking !’.

Even the roses in his garden did not please him anymore – the largest and sweetest and most beautiful roses ever seen – because they were not made of gold. And although the king was very fond of music in his youth, the only music he loved now was the sound of gold coins, one against another. At last, King Midas could not bear to touch anything that was not gold. He used to go down to a secret room under his palace where he kept his precious store. He would let himself in and count his gold pieces. He would hold the bars of gold and admire his gold cups and plates until he could hardly bear to leave them.

Now in those days, a great many wonderful things used to happen just as they do today. One morning King Midas was in his treasure room when he noticed that the sun was shining into the room more brightly than usual. Not only that, but a stranger stood there, smiling at him in the light of the sunbeam. King Midas knew that he had locked himself in as usual, and so he guessed that his visitor was no ordinary person. The stranger looked at the gold pieces that the king was counting. ‘You seem to be a very rich man’ he said. ‘But it has taken me a long time to collect this gold’, said King Midas. ‘If I could live a thousand years, I might have time to get richer.

Questions :
(i) Why did King Midas like being a king?
(ii) How did he respond, when his daughter came to him with a bunch of beautiful yellow flowers?
(iii) What did the king love deeply in his youth?
(iv) Why did he not want to touch anything at last?
(v) What led the king to guess that the stranger was not an ordinary person?

Answers :
(i) King Midas liked being a king, mainly because he was fond of his golden crown.
(ii) When his daughter came to him with a bunch of beautiful yellow flowers, he would pick them, if they were as golden as they looked.
(iii) The king loved music deeply in his youth.
(iv) His frenzied desire for gold refrained him from touching anything at last. In other words, he was not interested to touch anything that was not gold.
(v) In spite of his treasure room being locked by himself, to his astonishment the king found the stranger inside it. This led the king to guess that the stranger was not an ordinary person.

2. Read through the extract and answer the questions that follow.
The next morning, King Midas awoke before dawn. He looked eagerly to see if his bed had been turned into gold. But no; it was exactly as it had been before. He lay, very disappointed, looking around his room. Suddenly, the earliest sunbeam of the rising sun shone through the window and up to the ceiling above. It seemed to reflect its golden light toward him. Looking at the sheet on his bed, Midas was astonished to find that it had become cloth of gold. The Golden Touch had truly come to him, with the first sunbeam. King Midas got out of bed in excitement. He touched one of the legs of the bed as he did so – and it immediately became a golden pillar.

He pulled the curtain at the window, and at once it became golden, too. He put on his clothes and found himself dressed in golden cloth. He took up his spectacles and put them on – and he found he could see nothing at all. The glasses had turned into gold and he could not see through them. He took them off again. ‘Never mind’, he thought to himself. ‘The Golden Touch is worth more than a pair of spectacles, and Marygold will be able to read to me.’ King Midas went downstairs and into the garden. He noticed that even the brass handle of the door became gold as soon as he turned it. Then he went among the rose trees that had always been his pride and joy in the past.

When he went to breakfast that morning, he felt more hungry than usual. While he was waiting for his eggs to be ready, little Marygold came in crying bitterly. ‘Look, father !’ she cried, holding out a golden rose. ‘I went to pick you some roses and they are yellow and hard, and their sweet scent is gone.’ ‘Never mind, my dear’, said her father. ‘They are worth much more like that. Sit down and eat your breakfast.’ He poured himself a cup of coffee as he spoke. The coffee pot was a golden one when he put it back on the table. Then he tried a spoonful of coffee, to see if it was sweet enough. But it had become liquid gold.

Questions :
(i) When did King Midas get up the following morning?
(ii) How did he first make use of the Golden touch and what was the result?
(iii) What had always been his object of pride and happiness in the past?
(iv) ‘But it had become liquid gold.’ What does ‘it’ refer to?
(v) Suggest a suitable title for the extract.

Answers :
(i) The following morning, King Midas got up before the crack of dawn.
(ii) He made use of the Golden touch for the first time by touching one of the legs of the bed and at once it turned into a golden pillar.
(iii) The rose trees in his garden had always been the object of his pride and happiness in the past.
(iv) ‘It’ refers to a spoonful of coffee.
(v) King Midas and his irresistible temptation for gold.

3. Read through the extract and answer the questions that follow.
In despair, Midas looked about him. Suddenly he saw the stranger that had visited him the day before. ‘Well Midas’, said the stranger. ‘How do you like having the Golden touch ?’ ‘I have lost everything I really loved’, said King Midas. ‘I am full of sorrow and regret. Gold is of no use to me now.’ ‘So you have learnt something since yesterday ?’ asked the stranger. ‘Now which is worth more – the Golden Touch or a cup of cold water ?’ ‘Oh, blessed water !’ exclaimed Midas. ‘Will I ever taste it again?’ ‘The Golden Touch – or a piece of bread ?’ the stranger said. ‘A piece of bread’, answered Midas, ‘is worth all the gold on earth !’

‘Gold – or your own little daughter ?’ asked the stranger. ‘Oh – my child, my child !’ cried poor Midas. ‘I would not have given one hair of her head for the power to change the whole earth into gold !’ The stranger looked seriously at King Midas. ‘You are wiser than you were’, he said. ‘Your heart is still flesh and blood. You know truly that the common things of life, which are within everyone’s reach, are more valuable than riches. Tell me, do you want to keep the Golden Touch ?’ ‘No, it is hateful to me now’, said Midas, passionately. A fly settled on the king’s nose and immediately fell to the floor, a small scrap of gold.

Midas shuddered. ‘Then go down to the end of your garden’, said the stranger, ‘and wash in the water of the river there. Then bring some of the same water and sprinkle it over anything that you wish to change back again. If you do this, truly and sincerely, you can set right again the results of your greed of gold.’ King Midas bowed his head. When he looked up again, the stranger had vanished. The king ran at once to the river. Without waiting to take off his clothes, he dived in. In the coolness of the water, he felt at once that a weight had been lifted from his heart and body.

Questions :
(i) What was the king’s response to the stranger’s question concerning the possession of the Golden Touch?
(ii) “Well I ever taste it again ?” What does ‘it’ refer to?
(iii) Why did Midas tremble in fear?
(iv) What were the results of the king’s greed of gold?
(v) ‘The felt at once that a weight had been lifted from his heart and body.’ What does the ‘weight’ refer to?

Answers :
(i) The king’s response to the stranger’s question concerning the possession of the Golden Touch was only deep sorrow and regret. He was sad at the loss of everything he loved.
(ii) ‘It’ refers to a cup of cold water the king yearned for.
(iii) Midas trembled in fear at the sight of a fly that perched on his nose and at once fell to the floor, eventually turning into a small scrap of gold.
(iv) As a result, of his greed for gold, the king lost everything including precious water and his beloved little daughter Marygold.
(v) The ‘weight’ refers to Midas’s boundless greed for gold.

II. Multiple Choice Questions (MCQs) with Answers
Choose the correct option.

Unit – I
The text
Long ago ……… stranger had go.

Question 1.
Long long ago there lived a very rich man called :
(a) Midas
(b) Devdas
(c) Raidas
(d) Bidas
Answer:
(a) Midas

Question 2.
Besides being rich, Midas was a :
(a) merchant
(b) Minister
(c) Chief
(d) king
Answer:
(d) king

Question 3.
Midas had a little daughter called :
(a) Marygold
(b) Rose
(c) Sunshine
(d) Nainegold
Answer:
(a) Marygold

Question 4.
King Midas liked being a king, chiefly because he loved his :
(a) golden chair
(b) golden crown
(c) people
(d) name and fame
Answer:
(b) golden crown

Question 5.
King Midas was a lover of more than anything else in the world.
(a) Silver
(b) platinum
(c) gold
(d) Diamond
Answer:
(c) gold

Question 6.
King Midas wished everything to turn into :
(a) real platinum
(b) real diamond
(c) real silver
(d) real gold
Answer:
(d) real gold

Question 7.
The king was very fond of music in his youth, the only music he loved now was :
(a) the sound of birds
(b) the sound of gold coins
(c) the roar of wild animals
(d) the sound of drums
Answer:
(b) the sound of gold coins

Question 8.
King Midas could not bear to touch anything that was not:
(a) plastic
(b) Silver
(c) gold
(d) Diamond
Answer:
(c) gold

Question 9.
King used to go down to a secret room under his palace where he kept his :
(a) precious gold
(b) coal
(c) precious metals
(d) valuable books and maps
Answer:
(a) precious gold

Question 10.
One morning King Midas was in his treasure room and he noticed that:
(a) the sun was shining into the room more brightly than usual
(b) birds were singing a song beautifully
(c) the golds were being doubled magically
(d) the sun had not risen yet
Answer:
(a) the sun was shining into the room more brightly than usual

Question 11.
Who do you think, standing in front and smiling at the king in the light of a sunbeam
(a) king’s daughter Marygold
(b) the queen
(c) a stranger
(d) the minister
Answer:
(c) a stranger

Question 12.
King Midas knew that he had locked himself in as usual and so he guessed that his visitor was :
(a) no special person
(b) no ordinary person
(c) no poor person
(d) no rich person
Answer:
(b) no ordinary person

Question 13.
Midas thought carefully. This was a wonderful chance, and he felt that the stranger had :
(a) spiritual powers
(b) magical powers
(c) no power
(d) physical powers
Answer:
(b) magical powers

Question 14.
“I am tired of collecting my riches so slowly. I wish everything I touch could be turned into gold.” Who said this?
(a) daughter Marygold
(b) the minister
(c) the stranger
(d) the king
Answer:
(d) the king

Question 15.
The stranger granted his wish to be fulfilled, i.e.
(a) wish to have a son
(b) the Golden Touch
(c) three wishes
(d) none of the above
Answer:
(b) the Golden Touch

Unit – II
The text
The next morning ……..what had be done?

Question 16.
The next morning, King Midas awoke before dawn and looked eagerly to see :
(a) if his bed had been turned into gold
(b) if his bed had been turned into a bed of roses
(c) if his bed had been turned into a hanging swing
(d) none of these
Answer:
(a) if his bed had been turned into gold

Question 17.
The Golden Touch had truly come to the king :
(a) with his first touch
(b) with the first sunbeam
(c) with his first sight
(d) all of the above
Answer:
(b) with the first sunbeam

Question 18.
After a touch, the things turn into gold. And with this he found himself in :
(a) a sad mood
(b) an angry mood
(c) excitement
(d) a worrying situation
Answer:
(c) excitement

Question 19.
Marygold was holding out a golden rose.
(a) happy
(b) angry
(c) sad
(d) crying bitterly
Answer:
(d) crying bitterly

Question 20.
King Midas tried a spoonful of coffee, to see if it was sweet enough. But it had become :
(a) poison
(b) sour
(c) liquid gold
(d) bitter
Answer:
(c) liquid gold

Question 21.
But the eggs, the fish, the bread, the butter, and all the food was uneatable for the
(a) daughter Marygold
(b) king
(c) stranger
(d) Queen
Answer:
(b) king

Question 22.
King Midas turned annoyed, sad, and worried because :
(a) he was unable to eat anything because of his Golden Touch
(b) he was unable to rule over his state
(c) he was unable to see anything
(d) none of the above
Answer:
(a) he was unable to eat anything because of his Golden Touch

Question 23.
Midas bent down and kissed his :
(a) gold coins
(b) little daughter
(c) cups and plates
(d) all of the above
Answer:
(b) little daughter

Question 24.
What do you think, that might happen to Marygold after getting a kiss from his father?
(a) she became more affectionate toward her father
(b) she became happy
(c) she turns into an ugly girl
(d) she became a statue of gold
Answer:
(d) she became a statue of gold

Question 25.
What terrible change came over Marygold? Her sweet little face, lovely hair, and little body turned into.
(a) yellow gold, golden metal, and a figure of soid gold
(b) white diamond, shinning metal, and a hard substance
(c) shining platinum and precious metal
(d) none of the above
Answer:
(a) yellow gold, golden metal, and a figure of sold gold

Unit – III
The text
This story ………..roses.

Question 26.
King Midas felt so sad and sorrowful that he wished, he was the in all the world, if only his beloved daughter could be herself again.
(a) richest man
(b) happiest man
(c) poorest man
(d) honest man
Answer:
(c) poorest man

Question 27.
In despair, Midas looked about him and suddenly he saw that had visited him the day before.
(a) stranger
(b) Marygold
(c) known person
(d) none of the above
Answer:
(a) stranger

Question 28.
“I have lost everything I really loved; I am full of sorrow and regret. Gold is of no use to me now.” What does the expression show?
(a) the king is in excitement
(b) the king is sad
(c) the king is repenting for his deed
(d) the king is happy, what happened
Answer:
(c) the king is repenting for his deed

Question 29.
After having the joy of ‘the Golden Touch’, the king’s view changed :
(a) the Golden Touch is worthful than anything
(b) the Golden Touch is of no use if man’s need is not satisfied
(c) both (a) and (b)
(d) none of the above
Answer:
(b) the Golden Touch is of no use if man’s need is not satisfied

Question 30.
Midas wanted everything back to normal because :
(a) those were worthful
(b) those were his wants
(c) those were useless
(d) those were his needs
Answer:
(d) those were his needs

Question 31.
“I wouldn’t have given one hair of her head for the power to change the whole earth into gold!” This expression said by the king shows :
(a) his hate for his daughter
(b) duty towards his daughter
(c) love for his daughter
(d) all of the above
Answer:
(c) love for his daughter

Question 32.
“You are wiser than you were,” he said. “Your heart is still flesh and blood.” Here ‘you’ and ‘he’ stands for
(a) king and daughter
(b) king and stranger
(c) stranger and king
(d) daughter and stranger
Answer:
(b) king and stranger

Question 33.
Word ‘shuddered’ means
(a) tremble or shake violently
(b) rearrange
(c) avoid or reject
(d) past part
Answer:
(a) tremble or shake violently

Introducing the Author:
Nathaniel Hawthorne is an American novelist and short story writer. Much of Hawthorne’s writing centers on New England, with many works featuring moral allegories with a puritanical inspiration. His fiction works are considered part of the Romantic movement and more specifically, dark romanticism. His themes often center on the inherent evil and sin of humanity, and his works have deep psychological complexity.

About the Topic:
‘The Golden Touch, as the title implies, deals with King Midas’s boundless greed for gold. The inevitable happened. The legendary king sank into despair. At last his obsession with the yellow metal filled him with great repentance and changed his attitude.

Summary:
Hawthorne’s story, ‘The Golden Touch’, takes us back to a long past when there lived king Midas who was vastly wealthy. He was the father of a little daughter Marygold by name. His fascination in gold was more than anything else in the world. He also loved his daughter deeply. The spectacle of the golden light of the sun evoked his strong wish – everything could change into real gold. Even the largest and sweetest and most beautiful roses paled into significance before this precious yellow metal. In his youth, he loved music deeply, but now the sound of gold coins, one against another fascinated him most.

At last, kind Midas’s desire for gold became irresistible. He became a frequent visitor to a secret treasure room under his palace. He would allow himself in, count his gold pieces and hold the bars of gold. One morning, when the sun was shining brightly, he was in his treasure room which was locked inside. In the meantime, he noticed an unknown person, standing there, giving a smile at him in the sunlight. His amazement knew no bounds to see the stranger. The king did not make out how he came inside the locked room. As a result, he guessed the stranger was not an ordinary person.

The unknown person came to know of the king’s insatiable desire for gold; therefore, the former asked the latter what he wanted. The king expressed his wish that everything he touched could transform into gold. His wish to have the Golden Touch filled the stranger with surprise. He asked the king if he would not regret it. The king’s response was swift – it would give him perfect happiness. The stranger fulfilled his wish. He said to the king that the next day at sunrise, he would have his coveted Golden touch. The light of the sunbeam was too bright for Midas to see anything around him.

To his amazement, the stranger had vanished, when he opened his eyes. The following morning, the king discovered that his desire for the golden touch had not been fulfilled. His bed had not changed into gold. His sadness did not last long. A sudden sight of the reflection of the golden light of the earliest sunrise on him worked wonders. The sheet on his bed turned into a cloth of gold. The first sunbeam had truly brought the Golden Touch. Driven by excitement, he touched everything – one of the legs of the bed, the curtain at the window, his dress, and spectacles.

There was gold everywhere. The loss of spectacles didn’t matter to him much. The gold was more precious than his pair of spectacles and his daughter Marygold would read to him. The magic touch turned everything into gold beginning from the brass handle of the door, and the rose trees, the constant source of his pride and joy in the past. At the moment, he went into breakfast that morning, his little beautiful daughter Marygold came in crying bitterly. When he kissed her, she wore a terrible look, with her little face, beautiful hair, and her little body gone.

She became a hard golden figure. We find king Midas in a dejected mood. He plunged into deepening despair at the loss of everything he loved including his dear little daughter Marygold. In the meantime, the stranger reappeared and enquired him about his Golden Touch. He narrated his tale of woe to the former. He now realized the futility of the Golden Touch. He was terribly thirsty and pined for a cup of cold water to taste. The stranger kept on asking him what he preferred – the Golden touch or a piece of bread and gold or his own little daughter.

The grief-stricken father wanted his daughter to get back. His repentance made the stranger remark that he was wiser than before. The stranger knew human nature dies hard and the king was no different. Midas had now become a virulent hater of gold. He trembled in fear at the sight of a fly perching on his nose at once felt the ground turning into a piece of gold. In order to get rid of the burden of his Golden Touch, the stranger advised him to go down to the end of his garden, wash in the water of the river there, bring some of the same water and sprinkle it over anything he wished to change back again.

If the king does this sincerely and truly he can rectify his greed for gold. The king magnificently rose to the occasion. The story ends on a happy note. At first, the king sprinkled the water on the golden figure of his little daughter, Marygold. The inevitable happened. He got back his daughter again. Marygold was kept in dark about this unfortunate and painful incident. The king and his daughter lived happily.

ସାରାଂଶ:
ହଥର୍ୟଙ୍କ ଗଳ୍ପ ‘The golden Touch’ ଆମକୁ ଭସାଇଦିଏ ଏକ ସୁଦୂର ଅତୀତକୁ ଯେତେବେଳେ ଅହେତୁକଭାବେ ସୁନା ପ୍ରତି ଲୋଭ ଥିବା ଏକ ଶକ୍ତିଶାଳୀ ଓ ଧନୀ ରାଜା ବାସ କରୁଥିଲେ ଯାହାଙ୍କ ନାମ ଥିଲା ମିଦାସ୍ । Marygold ନାମକ ତାଙ୍କର ଗୋଟିଏ କୁନି ଝିଅ ଥିଲା। ପୃଥ‌ିବୀର ସବୁ ଜିନିଷଠାରୁ ସେ ସୁନାକୁ ବେଶି ଭଲ ପାଉଥିଲେ । ତା’ ସହିତ ସେ ତାଙ୍କର କନ୍ୟାକୁ ମଧ୍ୟ ଅତି ନିବିଡ଼ଭାବେ ଭଲ ପାଉଥିଲେ । ଅସ୍ଥାୟମାନ ସୂର୍ଯ୍ୟଙ୍କର ସୁନେଲି କିରଣକୁ ଦେଖ୍ ସେ ଭାବୁଥିଲେ ଏହା ସବୁ ଜିନିଷକୁ ସୁନାରେ ପରିଣତ କରିପାରନ୍ତା କି ? ତାଙ୍କ ଦୃଷ୍ଟିରେ ଏହି ମୂଲ୍ୟବାନ୍ ସୁନେଲୀ ଧାତୁର ମୂଲ୍ୟତୁଳନାରେ ସୁନ୍ଦର ଗୋଲାପଗୁଡ଼ିକର ମୂଲ୍ୟ କିଛି ନ ଥିଲା । ତାଙ୍କ ଝିଅ ଆଣିଥିବା ସୁନେଲି ରଙ୍ଗର ଫୁଲଟିକୁ ଦେଖି ରାଜା ଖୁସି ହୋଇ ନ ଥିଲେ କାରଣ ତାହା ସୁନାରେ ନିର୍ମିତ ନ ଥିଲା ।

ତାଙ୍କ ଯୁବାବସ୍ଥାରେ ସେ ସଙ୍ଗୀତକୁ ଗଭୀରଭାବେ ଭଲ ପାଉଥିଲେ ଏବଂ ଏବେ ଭଲ ପାଉଛନ୍ତି ସୁନାକୁ । ପରିଶେଷରେ ସୁନା ପ୍ରତି ତାଙ୍କର ଅହେତୁକ ଲୋଭ ବୃଦ୍ଧି ପାଇବାରେ ଲାଗିଲା । ତାଙ୍କ ପ୍ରାସାଦରେ ଥ‌ିବା ମୂଲ୍ୟବାନ ଜିନିଷର ଭଣ୍ଡାର ଏକ ଗୋପନ କୋଠରିକୁ ସେ ସୁନାର ମୁଦ୍ରା ଗଣିବାକୁ ବାରମ୍ବାର ପ୍ରବେଶ କରୁଥିଲେ । ତାଙ୍କ ପାଖରେ ସୁନାର ସ୍ତମ୍ଭ, ସୁନାର ପ୍ଲେଟ୍ ଓ ସୁନାର ପାଣିପାତ୍ର ସବୁ ରହିଥିଲା । ଦିନେ ସକାଳେ ଯେତେବେଳେ ସୂର୍ଯ୍ୟଙ୍କର ସୁନେଲି କିରଣ ବିଛେଇ ହୋଇ ପଡ଼ିଥିଲା, ଭିତର ପାଖରୁ ବନ୍ଦଥ‌ିବା ଗୁପ୍ତ କୋଠରି ଭିତରେ ଥାଇ ସେ ଦେଖିଲେ ଯେ ଜଣେ ଅପରିଚିତ ବ୍ୟକ୍ତ ସେଠାରେ ଠିଆ ହୋଇ ତାଙ୍କୁ ଚାହିଁ ସ୍ମିତହାସ୍ୟ କରୁଛନ୍ତି । ରାଜା ଅନୁମାନ କଲେ ଯେ ସେହି ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ଜଣେ ସାଧାରଣ ମଣିଷ ହୋଇ ନପାରନ୍ତି । ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ରାଜାଙ୍କର ସୁନା ପ୍ରତି ଥିବା ଅତୃପ୍ତ ଲୋଭ ବିଷୟରେ ଜାଣିପାରିଲେ ।

ରାଜା କ’ଣ ଚାହାନ୍ତି ବୋଲି ସେ ପ୍ରଶ୍ନ କଲେ । ରାଜା ଜାଣିଥିଲେ ସେ ଜଣେ ଅସାଧାରଣ ଅଲୌକିକ ବ୍ୟକ୍ତି ଥିଲେ । ତେଣୁ ରାଜା ବର ପ୍ରାର୍ଥନା କଲେ ଯେ ସେ ଯାହା ଛୁଇଁବେ ସେସବୁ ସୁନାରେ ପରିଣତ ହୋଇଯାଉ । ରାଜାଙ୍କର ଏହି ପ୍ରାର୍ଥନାରେ ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣକ ଆଶ୍ଚର୍ଯ୍ୟ ହୋଇଗଲେ ଏବଂ ରାଜାଙ୍କୁ ପଚାରିଲେ ତାଙ୍କର ଏଥ‌ିରେ କୌଣସି ଅନୁଶୋଚନା ରହିବ ନାହିଁ ତ ! ରାଜା ଚଳଚଞ୍ଚଳ ମନରେ ଉତ୍ତର ଦେଲେ ଯେ ସେ ଏହି ବର ପାଇଲେ ଅତ୍ୟନ୍ତ ଖୁସି ହେବେ। ରାଜାଙ୍କର ଇଚ୍ଛା ପୂର୍ଣ ହେଉ ବୋଲି ସେ ବର ଦେଲେ । ସେ ରାଜାଙ୍କୁ କହିଲେ ଯେ ତା’ ପରଦିନ ସୂର୍ଯ୍ୟୋଦୟ ସମୟରେ ସେ ଏହି ସ୍ଵର୍ଣ୍ଣ ସ୍ପର୍ଶର ଫଳ ପ୍ରାପ୍ତ ହେବେ । ସୂର୍ଯ୍ୟଙ୍କ ରଶ୍ମି ଏତେ ଉଜ୍ଜ୍ଵଳ ଥିଲା ଯେ ରାଜା ତାଙ୍କ ଚତୁପାର୍ଶ୍ଵରେ ଥ‌ିବା କୌଣସି ଜିନିଷ ଦେଖି ପାରୁ ନ ଥିଲେ । ଯେତେବେଳେ ରାଜା ଆଖି ଖୋଲିଲେ, ସେ ସେହି ଅଦୃଶ୍ୟ ବ୍ୟକ୍ତିଜଣଙ୍କୁ ଦେଖିବାକୁ ପାଇଲେ ନାହିଁ ।

ସେ ଅଦୃଶ୍ୟ ହୋଇ ଯାଇଥିଲେ । ପରଦିନ ରାଜା ଶୀଘ୍ର ଶଯ୍ୟାତ୍ୟାଗ କରି ଦେଖିଲେ ଯେ ତାଙ୍କର ସ୍ଵର୍ଣ୍ଣ ସ୍ପର୍ଶର ବରଟି ପରିପୂର୍ଣ୍ଣ ହୋଇନାହିଁ । ତାଙ୍କର ଶଯ୍ୟା ସୁବର୍ଣ୍ଣରେ ପରିଣତ ହୋଇନାହିଁ । ସେ ଦୁଃଖରେ ଭାଙ୍ଗିପଡ଼ିଲେ । ତା’ପରେ ସୂର୍ଯ୍ୟଙ୍କର ସୁନେଲି କିରଣ ଝରକା ଦେଇ ତାଙ୍କ ଶଯ୍ୟାରେ ପଡ଼ିଲା । ତାଙ୍କ ବିଛଣା ଚାଦରଟି ସୁନାରେ ପରିଣତ ହୋଇଥିବା ଦେଖ୍ ସେ ଖୁସିରେ ଆତ୍ମହରା ହୋଇଗଲେ । ସେ ସବୁ ଜିନିଷକୁ ସ୍ପର୍ଶ କରିବାକୁ ଇଚ୍ଛାକଲେ । ତା’ପରେ ଖଟର ଗୋଟେ ଗୋଡ଼, ଝରକାର ପରଦା ଓ ନିଜ ପୋଷାକକୁ ଛୁଇଁଦେଲେ, ସବୁଯାକ ଜିନିଷ ସୁନା ପାଲଟିଗଲା । ତାଙ୍କର ସ୍ପର୍ଶରେ ତାଙ୍କର ଚଷମାଟି ମଧ୍ଯ ସୁନା ପାଲଟିଗଲା । ଏଥରେ ତାଙ୍କର ତିଳେମାତ୍ର ମନଦୁଃଖ ହେଲା ନାହିଁ । କାରଣ ଚଷମା ଅପେକ୍ଷା ସୁନା ଥିଲା ତାଙ୍କ ପାଇଁ ଅଧିକ ମୂଲ୍ୟବାନ୍ । ଏହି ଚମତ୍କାର ସ୍ପର୍ଶରେ ସବୁକିଛି ସୁନାରେ ପରିଣତ ହୋଇଗଲା ।

ତାଙ୍କ ଅତୀତର ଗର୍ବ ଓ ଖୁସିର ଉତ୍ସ ଗୋଲାପ ଗଛଗୁଡ଼ିକ ମଧ୍ୟ ସୁନାରେ ପରିଣତ ହୋଇଗଲା । ଏହି ସମୟରେ ତାଙ୍କ ପାଖକୁ ତାଙ୍କର ଝିଅ ଏକ ସୁନାର ଗୋଲାପ ଫୁଲ ଧରି କାନ୍ଦି କାନ୍ଦି ଆସିଲା ଏବଂ ଏହାର ବାସ୍ନା ଚାଲିଯାଇଥିବାରୁ ବ୍ୟସ୍ତ ହୋଇପଡ଼ିଲା । ରାଜା ନିଜେ କଫି ପିଇବାକୁ ଇଚ୍ଛା କରି, ସେ କଫି ଗ୍ଲାସ୍‌କୁ ସ୍ପର୍ଶ କରନ୍ତେ ତାହା ତରଳ ସୁନା ପାଲଟିଗଲା । ଭୋକିଲା ରାଜା ଜଳଖୁଆ ପାଇଁ ମାଛ, ଅଣ୍ଡା, ରୁଟି ଓ ଲହୁଣି ଖାଇବାକୁ ଚାହାନ୍ତେ, ସେସବୁ ତାଙ୍କ ସ୍ପର୍ଶ ପାଇବା ପରେ ସୁନାରେ ପରିଣତ ହୋଇଗଲା । ସେ ଖାଇ ନ ପାରିବାରୁ ମନ ଦୁଃଖ କଲେ । ଏହି ସମୟରେ ତାଙ୍କ ମନଦୁଃଖର କାରଣ ବିଷୟରେ ତାଙ୍କ କୁନି ଝିଅ ତାଙ୍କୁ ପଚାରିଲେ । ରାଜା ତାଙ୍କୁ ଚୁମ୍ବନ ଦେଲେ । ଫଳରେ ମେରିଗୋଲ୍ଡର ଶରୀରରେ ଏକ ଭୟଙ୍କର ପରିବର୍ତ୍ତନ ଘଟିଲା । ତାଙ୍କର କୁନି ସୁନ୍ଦର ମୁଖମଣ୍ଡଳ, ସୁନ୍ଦର କେଶରାଶି ଏବଂ କୁନି କୋମଳ ଶରୀରଟି କଠିନ ସୁନାରେ ପରିଣତ ହୋଇଯାଇଛି ।

ଯେତେବେଳେ Marygoldର ଶରୀରଟି ସୁନା ପାଲଟିଗଲା, ରାଜା ଆଶ୍ଚର୍ଯ୍ୟ ହୋଇଗଲେ ଏବଂ ଦୁଃଖ ଓ ଶୋକରେ ଭାଙ୍ଗିପଡ଼ିଲେ । ସବୁ ଜିନିଷ ସୁନାରେ ପରିଣତ ହୋଇଯିବା ହେତୁ ସେ ଗଭୀର ଦୁଃଖରେ ମର୍ମାହତ ହୋଇଗଲେ । ନିଜ ଝିଅର ପୂର୍ବ ଅବସ୍ଥା ଫେରି ପାଇବାପାଇଁ ସେ ବ୍ୟାକୁଳ ହୋଇ ଉଠିଲେ । ଏହି ସମୟରେ ସେହି ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣକ ପୁନର୍ବାର ଆବିର୍ଭାବ ହେଲେ ଏବଂ ସେହି ବରର ପୂର୍ଣ୍ଣତା ବିଷୟରେ ଜାଣିବାକୁ ଚାହିଁଲେ । ରାଜା ତାଙ୍କ ଦୁଃଖର କାହାଣୀ ବର୍ଣ୍ଣନା କଲେ । ଏହି Golden Touchର ମୂଲ୍ୟହୀନତା ବିଷୟରେ ସେ ଅନୁଭବ କରିପାରିଲେ । ସେହି ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ରାଜାଙ୍କୁ ପଚାରିଲେ ସୁନା କିମ୍ବା ଏକ ଗ୍ଲାସ୍ ପାଣି କେଉଁଟି ବିଶେଷ ଦରକାର ? ରାଜା ଉତ୍ତର ଦେଲେ ଜୀବନରେ ବଞ୍ଚିବାପାଇଁ ପାଣିର ଆବଶ୍ୟକତା ଅପରିହାର୍ଯ୍ୟ ।

ଅପରିଚିତ ବ୍ୟକ୍ତିଟି ପୁନର୍ବାର ରାଜାଙ୍କୁ ପଚାରିଲେ ସୁନା ଏବଂ ରୁଟି ମଧ୍ୟରୁ ତାଙ୍କ ପାଇଁ କେଉଁଟି ଅଧ୍ଵ ମୂଲ୍ୟବାନ୍ ? ତା’ପରେ ପଚାରିଲେ ସୁନା କିମ୍ବା ତାଙ୍କର କୁନି ଝିଅ ମଧ୍ୟରୁ ତାଙ୍କ ପାଇଁ କେଉଁଟି ମୂଲ୍ୟବାନ ? ଦୁଃଖରେ ଭାଙ୍ଗିପଡ଼ିଥିବା ରାଜା ତାଙ୍କ ଝିଅକୁ ପୁନର୍ବାର ଫେରି ପାଇବାପାଇଁ ବ୍ୟାକୁଳ ହୋଇଉଠିଲେ । ରାଜାଙ୍କର ଶୋଚନାରୁ ଜଣାଗଲା ଯେ ରାଜା ପୂର୍ବାପେକ୍ଷା ଅଧିକା ଜ୍ଞାନୀ ହୋଇ ପାରିଛନ୍ତି । ରାଜା ସୁନାକୁ ଘୃଣା କରୁଥିଲେ । ତାଙ୍କ ନାକରେ ବସିଥିବା ମାଛିଟି ତତ୍‌କ୍ଷଣାତ୍ କ୍ଷୁଦ୍ର ସୁନାଖଣ୍ଡଟିଏ ହୋଇ ମାଟିରେ ପଡ଼ିଲା । ଏହି ସ୍ପର୍ଷ ସ୍ପର୍ଶ ବରପ୍ରାପ୍ତିରୁ ମୁକ୍ତ ହେବା ପାଇଁ ରାଜା ବ୍ୟାକୁଳ ହୋଇପଡ଼ିଲେ । ଅପରିଚିତ ବ୍ୟକ୍ତିଜଣଙ୍କ ତାଙ୍କୁ ଉପଦେଶ ଦେଇ କହିଲେ ଯେ ବଗିଚାର ଶେଷରେ ଥିବା ନଦୀକୁ ଯାଇ ସ୍ନାନ କରିବେ ଏବଂ କିଛି ଜଳ ଆଣି ଯାହାକୁ ପୂର୍ବାବସ୍ଥାକୁ ଆଣିବାକୁ ଚାହୁଁଥ‌ିବେ ତାହାକୁ ତା’ ଉପରେ ସିଞ୍ଚନ କରିବେ ।

ତା’ପରେ ସେ ସେସବୁକୁ ତା’ର ପୂର୍ବ ଅବସ୍ଥାରେ ଫେରି ପାଇବେ । ସେ ତତ୍‌କ୍ଷଣାତ୍‌ ନଦୀରେ ସ୍ନାନ କଲେ ଓ ନଦୀରୁ ପାଣି ଆଣି ନିଜ ଝିଅ ଉପରେ ସିଞ୍ଚନ କଲେ ଓ ଅନ୍ୟାନ୍ୟ ଜିନିଷ ଉପରେ ମଧ୍ୟ ସିଞ୍ଚନ କଲେ । ଫଳରେ ସେଗୁଡ଼ିକ ପୂର୍ବ ଅବସ୍ଥାକୁ ଫେରିଆସିଲା। Marygoldଙ୍କୁ ଏହି ଦୁର୍ଭାଗ୍ୟ ଓ ଦୁଃଖଦ ଘଟଣା ବିଷୟରେ ଜଣାଇ ଦିଆଗଲା ନାହିଁ । ରାଜା ଅନୁଭବ କରିପାରିଲେ ଯେ ସୁନାର ସମୁଦ୍ର ଅପେକ୍ଷା ସେ ଆଣିଥିବା ପାଣି ମାଠିଆର ମୂଲ୍ୟ ବହୁତ ଅଧିକ । ଏହାପରେ ରାଜା ତାଙ୍କ ଝିଅକୁ ନେଇ ଖୁସିରେ ଜୀବନ ବିତାଇଲେ ।

Odisha State Board CHSE Odisha Class 11 Invitation to English 1 Solutions Chapter 5 The Cancer Fight, from Hiroshima to Houston Textbook Exercise Questions and Answers.

CHSE Odisha 11th Class English Solutions Chapter 5 The Cancer Fight, from Hiroshima to Houston

CHSE Odisha Class 11 English The Cancer Fight, from Hiroshima to Houston Text Book Questions and Answers

UNIT -I
Gist with Glossary:

Gist:
The writer takes us to her native Hiroshima in 1945 when it was completely destroyed by the atomic bomb explosion. Its long-term effects resulted in the deaths of about half of her relatives. Her father was also not spared. This tragedy evoked both her fascination with and her dread of radiation. The early death of her very dear friend Sadako Sasaki as a result of radiation-linked blood cancer or bone marrow made her take an oath to become a cancer doctor.

Today Dr. Komaki as the clinical chief and Program Director of Thoracic Radiation Oncology at The University of Texas M.D. Anderson Cancer Centre in Houston is one of the world’s distinguished researchers and advocates of proton radiation beam therapy with which radiation-related cancer patients are treated. Private hospitals are not suitable for the treatment of these patients on trial, because they lack adequate manpower, qualitative review boards, and maintenance of strict quality control of treatment. Komaki’s motto is to educate students, trainees, and patients to focus on living healthy lives. Besides, she wants to inspire them to reach their goal to help others.

Glossary:
exploded : shattered violently (ବିସ୍ଫୋରଣ ହେଲା)
devastated : completely destroyed (ସମ୍ପୂର୍ଣ୍ଣରୂପେ ଧ୍ବଂସ କରିଦେଲା)
fascination : charm (ଆକର୍ଷଣ)
leukemia : a type of cancer of the blood or bone marrow (ଏକପ୍ରକାର କର୍କଟ ରୋଗ)
sophisticated: advanced (ଆଧୁନିକ)
oncologist : a doctor who treats cancer (କର୍କଟ ରୋଗ ଚିକିତ୍ସକ)
review : a formal assessment of something to institute necessary changes (ତର୍ଜମା)
compliance : action according to standard (ମାନାନୁସାରୀ କାର୍ଯ୍ୟ)
innovators : those who introduces new thoughts (ନବ ପ୍ରଦର୍ଶକ)
achieve : fulfil (ହାସଲ କରିବା)
goal : objective (ଲକ୍ଷ୍ୟ)

Think it out:
Question 1.
How did Hiroshima become a part of world history?
Answer:
Hiroshima became a part of world history during the final stages of World War II in 1945 when America dropped the atomic bomb on Hiroshima. Many lives were lost. Innumerable people suffered from cancer of the blood or bone marrow. The city was completely destroyed.

Question 2.
What are the three traumatic events of Ritsuko’s childhood?
Answer:
The three traumatic events of Ritsuko’s childhood were the atomic bomb explosion on her native city Hiroshima in 1945, the deaths of about half her relatives, including her father and the death of her most intimate friend Sadako Sasaki at the age as a result of radiation-related blood cancer.

Question 3.
What were the results of these experiences for Ritsuko?
Answer:
The results of these experiences for Ritsuko were her vow to become a cancer doctor. Her dream came true and she became a famous cancer specialist.

Question 4.
What is Dr. Komaki’s specialization?
Answer:
Dr. Komaki’s specialization was the application of radiation creatively in the treatment of cancer.

Question 5.
What is she famous for?
Answer:
She is famous as one of the world’s leading researchers and advocates of proton radiation beam therapy.

Question 6.
How do oncologists view proton radiation beam therapy?
Answer:
Oncologists view proton radiation beam therapy as the safest and most effective.

Question 7.
Why does she prefer to work at a university?
Answer:
She prefers to work in a university because she can do her work much better in the backdrop of a university program that is not available at private institutions.

Question 8.
Do you think she likes to teach? Why do you think so?
Answer:
Yes, she likes to teach, because she is interested to share her knowledge with future scientists who introduce new ideas of things.

Question 9.
What is she more interested in research and patient care or money? Why do you think so?
Answer:
She is more interested in research and patient care than money. Her focus on proton radiation therapy to treat cancer patients, her keen interest in sharing her knowledge with future innovators, and her desire to create awareness among all sorts of people concerning healthy lives splendidly exemplify this fact.

Question 10.
What is her mission in life?
Answer:
Her mission in life is to educate younger people – students, trainees, and patients to lead healthy lives and achieve their objective to help others.

Question 11.
Explain the meaning of this expression : (she) no longer fears it.
Answer:
The atomic bomb explosion in her native city was devastating. It resulted in the loss of innumerable lives including his relatives and father. They all died of radiation-related cancer. Komaki was shocked. Later, he learned the technique of applying radiation in an innovative manner and hence no longer fears it.

UNIT – II
Gist with Glossary:

Gist:
This unit begins with the definition of proton therapy. It is a form of treatment that damages and eventually destroys cancerous cells of the exact spot of the affected tumor with the help of charging particles. There is a difference between proton therapy and photon therapy; the latter is not as good as the former. Proton therapy is confined to a particular area. It is powerful too. Photon therapy, in spite of killing cancerous cells, has side effects. It leads to an increase in secondary cancer. Proton therapy plays a great role in striking specific tumors precisely. It results in the minimization of injury to other organs.

Besides, photons (X-rays) emit 90 percent of their cancer-fighting energy, as they pierce the skin. But their utility is lost by 40 percent. Proton therapy is more effective for patients who suffer from prostate cancer. Proton therapy is used for those whose localized cancer has not affected the distant parts of the body. It has a great effect on children in particular. Proton therapy also helps the elderly, because their excessively weak skin cannot resist the power of radiation therapy. Komaki’s patients think that proton therapy is fabulous because they can tolerate it more easily than photon treatment.

Glossary :
ionizing : charging (ଚାର୍ଜିତ)
tumor : a swelling part of body caused by an abnormal growth of tissue
ultimately : eventually (ଘଟଣାକ୍ରମେ )
radiation therapy : treatment of cancer by using radiation (ବିକିରଣଦ୍ବାରା ଚିକିତ୍ସା)
dosage : medicinal dose (ଔଷଧର ମାତ୍ରା)
hits : affects (କ୍ଷତିଗ୍ରସ୍ତ କରେ)
toxicity : the quality of being poisonous (ବିଷାକ୍ତ)
reduce : minimize (କମାଇ ଦେବା)
specific : particular (ନିର୍ଦ୍ଦିଷ୍ଟ )
precisely : exactly (ନିର୍ଦ୍ଦିଷ୍ଟଭାବେ)
in addition : besides (ଏଥ୍ ସହିତ)
feasible : viable (ଫଳପ୍ରଦଭାବେ କାର୍ଯ୍ୟ କରିବାର ସାମର୍ଥ୍ୟ)
penetrate : go deep (ପ୍ରବେଶ କରେ)
further : any more (ପୁନର୍ବାର)
calibrated : measured (ପରିମାପ କରାଗଲେ )
efficacy : effectiveness (ସଫଳତା)
simultaneously : at the same time (ଏକସମୟରେ)
collateral damage : injury to other organs (ଅନ୍ୟାନ୍ୟ ଅଙ୍ଗର କ୍ଷତି)
withstand : resist (ସହ୍ୟ କରିବା)
stresses : emphasizes (ଜୋର୍ ଦେବା)
fabulous : very great (ଖୁବ୍ ବେଶି )
amazing : astonishing (ଆଶ୍ଚର୍ଯ୍ୟକର)
tolerate : bear (ସହ୍ୟ କରିବା)

Think it out:
Question 1.
What is proton therapy?
Answer:
Proton therapy refers to the treatment of cancer by focusing charging particles on the exact tumor-affected spot to damage and eventually destroy its cancerous cells.

Question 2.
What are the damaging side effects of photon therapy?
Answer:
The damaging side effects of photon therapy are pneumonitis, esophagitis, and poisonous bone marrow. It also contributes to the growth of secondary cancers.

Question 3.
What are the advantages of proton therapy over photon therapy?
Answer:
The advantages of proton therapy are better in comparison to photon therapy, because of its confinement only to the affected part of the body, no side effects, and reduction in the destruction of other organs.

Question 4.
Do you think proton therapy is more effective at certain stages of cancer?
Answer:
Yes, I think proton therapy is more effective for prostate cancer.

Question 5.
For what stage of cancer is proton therapy the most useful?
Answer:
Proton therapy is the most useful for those whose localized cancer has not affected the distant parts of the body.

Question 6.
For which age groups is proton therapy the best? Why?
Answer:
Proton therapy is the best both for children and the elderly, because, it causes little, if any, injury to the former’s still-growing tissues and organs surrounding the tumors. In the same way, it helps the elderly because tissues and organs surrounding their tumors are very weak to resist usual radiation treatment.

Question 7.
Why do her patients think proton therapy is fabulous?
Answer:
Komaki’s patients think that proton therapy is fabulous because this treatment itself makes them seldom sick. Besides, it is astonishing, they can tolerate it easily in comparison to photon treatment.

UNIT-III
Gist with Glossary

Gist:
In the course of her externship, internship, residency, and fellowship in Milwaukee, Komaki came to know the importance of radiation in the treatment of cancer. Besides, she learned that localized radiation therapy was not so harmful as chemotherapy. In 1985, she along with Cox set out to Newyork to work with Dr. Eric Hall, who was, at that time, a renowned global authority on the effects of the atomic bomb on human beings. In 1988, she spent years of research work in this field at UT M.D. Anderson. Proton therapy started in 1954.

The Harvard Cyclotron Laboratory in partnership with Massachusetts General Hospital started treating cancer patients. The necessary technology was very costly. Therefore, the treatment did not go beyond the physics research lab until 1990. The first hospital-based program commenced with the opening of the Proton Treatment Center at Loma Linda University Medical Center in southern California. Their keen interest in proton therapy and the relentless effects of having a similar unit at UT M.D. Anderson, Huston at last bore fruit.

Now there are many such centers established in America. Today Komaki and her colleagues treat thoracic malignancies, at least 10 to 15 patients every day. Proton therapy continues for about seven weeks. In Komaki’s view, they have a long way to go. She opines that proton therapy is very important so far as the future of cancer care is concerned. She is hopeful to make the patients more comfortable by destroying cancer cells, but not killing normal cells. They can lead a normal life during treatments. Dr. Ritsuko Komaki and her colleagues at UT M.D. Anderson will be a leader in this field in the future.

Glossary :
viable : possible (ସମ୍ଭାବ୍ୟ)
chemotherapy: treatment of cancer using chemical substances
scattered: spread throughout
leading : famous (ପ୍ରସିଦ୍ଧ ବା ବିଖ୍ୟାତ)
commence: start or begin (ଆରମ୍ଭ ହେଲା)
expensive : costly (ବ୍ୟୟବହୁଳ)
intrigued : interested to know more (ଅଧୁକ ଜାଣିବାକୁ ଆଗ୍ରହୀ)
begin to push: to try their best
paid off : rewarded (ପୁରସ୍କୃତ ହେଲେ)
dosimetrists: experts in the dosage of
radiation scanning: cause a beam to traverse across a surface
secondary malignancy: a cancer that arises in the background of another malignancy
at the forefront: in the lead (ମୁଖ୍ୟ, ଆଗୁଆ)

Think it out:
Question 1.
What different kinds of therapies were used for treating cancer before the use of proton therapy?
Answer:
Different kinds of therapies such as surgery, radiation, chemotherapy, and photon theory were used for treating cancer before proton therapy.

Question 2.
What did Komaki initially learn about treating cancer?
Answer:
Komaki initially learned that surgery was the only possible method for treating cancer.

Question 3.
What did she learn about cancer treatment in the USA?
Answer:
In the U.S.A., she learned how radiation therapy was meant for cancer treatment.

Question 4.
What two things did Komaki learn about radiation?
Answer:
During her stay in the U.S.A., she learned that localized radiation treatment was not so harmful as chemotherapy in the treatment of cancer. Besides, she was aware of an important thing – it was quite different from radiation therapy.

Question 5.
How did Komaki and her husband start proton therapy at Anderson Cancer Center?
Answer:
Proton therapy research in the Proton Treatment Center at Loma Linda University Medical Center in southern California greatly interested Komaki and her husband. They visited Loma Linda to try their best to open a similar unit at UT M.D. Anderson in Houston. Their efforts were rewarded in 2006. The $ 125 million Proton Therapy Centre came into existence paving the way for a complete range of proton treatments.

Question 6.
When did proton therapy first start?
Answer:
Proton therapy was first started in 1954.

Question 7.
What is Komaki’s opinion on proton therapy?
Answer:
In Komaki’s opinion, proton therapy is of great importance to the future of cancer care. The patients, she views, can lead a normal life during treatments.

Post-Reading Activities:

Doing with words :
We can know the meanings of words by looking up a dictionary and finding out how a word had been used in a text. In a dictionary, words come in alphabetical order. The main word is called the headword. However, we should try to guess the meanings of words first from the context. That is the best way to learn new words.
(a) Match each word with its definition. Go back to the text for clues.
(b) Then, put the headwords in alphabetical order.

Answer:
(a)

(b)

CHSE Odisha Class 11 English The Cancer Fight, from Hiroshima to Houston Important Questions and Answers

I. Short Answer Type Questions with Answers

1. Read through the extract and answer the questions that follow.
Dr. Ritsuko Komaki was living with her family near Osaka when the atomic bomb exploded on her native Hiroshima in 1945. But the family returned to the devastated city when she was four, and Komaki grew up a witness to the long-term effects, which likely contributed heavily to the death of about half her relatives, including her father. Like many Japanese, she developed both a fascination with and fear of radiation. When her close friend Sadako Sasaki died at age 11 of radiation-related leukemia, Komaki vowed to become a cancer doctor.

Today, Dr. Komaki has learned how to apply radiation creatively and no longer fears it; instead, as clinical section chief and Program Director of Thoracic Radiation Oncology at The University of Texas M. D. Anderson Cancer Center in Houston, she uses it in increasingly sophisticated ways to fight cancer. She is one of the world’s leading researchers and advocates of proton radiation beam therapy, an emerging treatment that many oncologists consider the safest and most effective available.

And according to Komaki, her work is much better done in the context of a university program than at private institutions. “Our patients, including all of our proton patients, are treated under clinical trials,” which monitor results on large numbers of patients, she points out. “It’s difficult to treat patients on trials at private hospitals or clinics. They don’t have enough manpower, and they don’t have review boards that can check on compliance of eligibility and maintain strict quality control of treatment.

This means patients get better care here.” Recently awarded the Juan A. del Regato Foundation Gold Medal for best educator and teacher, the higher education setting also allows Dr. Komaki to share her knowledge with future innovators. “I try always to educate younger people – students and trainees, as well as patients – to live healthy lives and achieve their goal to help others.”

Questions :
(i) Why did Komaki take an oath to become a cancer doctor?
(ii) Name the university Dr. Komaki is working at now as clinical section chief and Program Director of thoracic Radiation Oncology.
(iii) “……….. which monitor results on large numbers of patients ”. What does the word ‘which’ refer to?
(iv) Why has she been recently awarded the Juan A. del Regato Foundation Gold Medal?
(v) What helps Dr. Komaki share her knowledge with future innovators?

Answers :
(i) Komaki took an oath to become a cancer doctor because her intimate friend Sadako Sasaki died of radiation-related leukemia when she was only 11.
(ii) The university Dr. Komaki is working at now as clinical section chief and Program Director of Thoracic Radiation Oncology is The University of Texas.
(iii) The word ‘which’ refers to the clinical trials meant for the treatment of all radiation-linked cancer patients.
(iv) She has been recently awarded the Juan A. del Regato Foundation Gold Medal for best educator and teacher.
(v) The higher education background helps Dr. Komaki share her knowledge with future innovators.

2. Read through the extract and answer the questions that follow.
Proton therapy, like other forms of radiation, aims ionizing particles (in this case, protons) onto a target tumor to damage and ultimately destroy its cancerous cells. But proton beams are much more localized and powerful than the X-rays used in more established radiation therapies. In the latter, the dosage is big enough to kill the cancerous cells, but some radiation also hits the healthy cells around the tumor. This can cause such side effects as pneumonitis, esophagitis, and bone marrow toxicity, or lead to the growth of secondary cancers.

To reduce damage to healthy tissues by a scattered dose of low radiation, the oncologist requires sharply delineated radiation, and proton beams are shaped to almost perfectly match the specific tumor and aimed to strike it precisely. Collateral damage is thus minimal, making it feasible to hit cancer with much bigger doses. In addition, photons (X-rays) release up to 90 percent of their cancer-fighting energy as they penetrate the skin, and lose 30 percent of it by the time they reach the tumor.

Meaning their overall effectiveness is reduced by 40 percent; they also exit out the rear of the tumor to further damage healthy tissues behind it. By controlling the speed with which it is shot into the body, the proton beam is calibrated to be at 30 percent of its maximum efficacy near the skin level while gathering full strength when it actually reaches the tumor and it barely exits the body at all.

Questions :
(i) ‘In the latter, the dosage is big enough to kill the cancer cells ….’ What does ‘the latter’ refer to?
(ii) What can be instrumental in the growth of secondary cancers?
(iii) “…………. as they penetrate the skin ” What does ‘they’ refer to?
(iv) What happens when they go into the skin?
(v) Give a suitable title to the extract.

Answers :
(i) ‘The latter’ refers to the X-rays used in more well-known radiation therapies.
(ii) When X-rays are used in radiation therapy, some radiation is likely to hit the healthy cells around the tumor, they can be instrumental in the growth of secondary cancers.
(iii) ‘They’ refers to photons (X-rays) that release up to 90 percent of their cancer-fighting energy as they pierce the skin.
(iv) When they go into the skin, photons lose 30 percent of their cancer¬fighting energy by the time they reach the tumor.
(v) Proton therapy and its advantages.

3. Read through the extract and answer the questions that follow.
Komaki says the treatment is most recommended for those whose localized cancer has not spread to distant parts of the body. The success rate against prostate cancer, for example, is around 90 to 95 percent. But proton therapy succeeds against many of the 130 known forms of cancer. A recent study led by Komaki and her husband Dr. James Cox, Head of the division of Radiation Oncology at UT M. D. Anderson Cancer Center, shows that proton therapy, when used simultaneously with chemotherapy to treat lung cancer, causes significantly less damage to surrounding healthy cells than other forms of radiation.

It’s also particularly effective for children, because it causes little if any, collateral damage to their still-growing tissues and organs. Likewise, the elderly are strong candidates because tissues and organs surrounding their tumors are often too weak to withstand the more commonly employed radiation treatments. Dr. Komaki also stresses that the side effects are minimal, noting one patient who played a round of golf following each of his daily outpatient treatments. “Patients think this is fabulous,” she says. “You know why? They rarely get sick from proton treatment itself! ‘Are you sure you’re giving me the treatment ?’ they ask. It’s amazing how much proton treatment they can tolerate compared to photon treatment.”

Questions :
(i) For whom is the proton therapy recommended?
(ii) Who is Dr. James Cox?
(iii) “It’s also particularly effective for children.” What does ‘it’ refer to?
(iv) What does the expression ‘collateral damage’ mean?
(v) What, according to the cancer patients, is quite astonishing?

Answers :
(i) Proton therapy is recommended for those whose localized cancer has not affected other distant parts of the body.
(ii) Dr. James Cox is the Head of the division of Radiation Oncology at UT M. D. Anderson Cancer Center. He is the husband of Dr. Komaki.
(iii) ‘It’ refers to proton therapy.
(iv) The expression ‘collateral damage’ refers to/means an injury caused to the patient’s other organs, besides the affected one.
(v) According to cancer patients, it is quite astonishing how they can tolerate proton treatment so easily in comparison to photon treatment.

4. Read through the extract and answer the questions that follow.
When she entered medical school in Hiroshima, Komaki was taught that surgery was the only viable cancer cure. But in the 1970s – while doing her externship, internship, residency, and fellowship in Milwaukee – she began learning “how radiation could cure people, and that gave it a different meaning to me than just the atomic bomb,” she says.

She came to view localized radiation treatment as less harmful than chemotherapy and realized that it couldn’t be equated at all with the scattered, uncontrolled radiation to the whole body that comes with exposure to an atomic bomb. In 1985, she and Cox went to the Columbia Presbyterian Medical Center in New York to work with Dr. Eric Hall, then a leading international authority on the effects of the atomic bomb on humans. In 1988, she began putting her years of research to work at UT M.D. Anderson.

Though protons were discovered by Ernest Rutherford in 1919, proton therapy didn’t commence until 1954, at Berkeley nuclear physics labs. The Harvard Cyclotron Laboratory partnered with Massachusetts General Hospital to begin treating cancer patients in 1961. But necessary technology is so expensive that treatment remained confined to physics research labs until 1990. That’s when the Proton Treatment Center opened at Loma Linda University Medical Center in southern California to offer the first hospital-based program.

Questions :
(i) Where and when did Dr. Komaki pursue her externship, internship, residency, and fellowship?
(ii) How does a person’s entire body become a victim of scattered and uncontrolled radiation?
(iii) Who was Dr. Eric Hall?
(iv) When did Komaki begin to spend her years of research working at UT M. D. Anderson?
(v) Why did the treatment of cancer patients wait till 1990?

Answers :
(i) Dr. Komaki pursued her externship, internship, residency, and fellowship in Milwaukee in the 1970s.
(ii) A person’s entire body becomes a victim of scattered and uncontrolled radiation that comes with exposure to an atom bomb.
(iii) Dr. Eric Hall was a prominent global authority on the effects of the atomic bomb on human beings.
(iv) In 1988 Komaki began to spend her years of research work at UT M. D. Anderson.
(v) The treatment of cancer patients waited till 1990, because of costly necessary technology.

5. Read through the extract and answer the questions that follow.
Intrigued by the proton therapy research then available, Komaki and Cox visited Loma Linda early on and began to push for a similar unit at UT M. D. Anderson when they returned to Houston. Their efforts paid off in 2006 when the $ 125 million Proton Therapy Center opened its doors offering a complete range of proton treatments. It is the only such facility in the Southwest. (In addition to Houston and Loma Linda, the nation now has centers in Florida, Indiana, and Massachusetts, with several, more being planned.)

Today, Komaki and her colleagues treat thoracic malignancies – therapy that requires a team of seven doctors, several physicians and dosimetrists, and a couple of dozen technicians – in 10 to 15 patients daily, with four of them taking proton therapy. Like other radiation treatments, proton therapy runs for about seven weeks. In all, the Proton Therapy Center, with Dr. Andrew Lee as medical director, treats 75 to 80 patients daily. To Komaki, this is just the beginning. “Our physicists have already made a machine here that creates a scanning beam,” she says.

“We’re the only place in the world with a scanning beam, which means we can remove neutron contamination.” (Neutrons are created when protons enter the body; though it happens rarely they can cause secondary malignancy a decade or more later.) She also believes that proton therapy is extremely important to the future of cancer care. “The patients are already sick from cancer,” she says, “why make them get sicker from treatment? Now, we can make them more comfortable – killing cancer cells but not killing normal cells. Now, they can live a normal life while getting treatments.” Whatever happens next in this field, Dr. Ritsuko Komaki and her colleagues at UT M. D. Anderson are likely to be at the forefront.

Questions :
(i) ‘………… and began to push for a similar unit at UT M. D. Anderson’ -What does ‘a similar unit’ refer to?
(ii) How did Komaki and Cox succeed in their efforts?
(iii) Name the places where a complete range of proton treatments is available.
(iv) Who is a dosimetrist ?
(v) How long does proton therapy continue?

Answers :
(i) ‘A similar unit’ refers to the proton therapy center.
(ii) Komaki and Cox succeeded in their efforts in 2006 when the $ 125 million Proton Therapy Centre was set up in Houston providing all sorts of proton treatments.
(iii) The places where a complete range of proton treatments is available are California, Houston, Florida, Indiana, and Massachusetts.
(iv) A dosimetrist is an expert in the dosage of radiation.
(v) Proton therapy continues for about seven weeks.

II. Multiple Choice Questions (MCQs) with Answers
Choose the correct option.

Unit – I
The text
Dr. Ritsuko Komaki ……….. to help others.

Question 1.
Name the author of the prose “The Cancer Fight, From Hiroshima to Houston”.
(a) Ritsuko Komaki
(b) Sadako Sasaki
(c) James Cox
(d) Eric Hall
Answer:
(a) Ritsuko Komaki

Question 2.
Dr. Ritsuko Komaki is a radition :
(a) oncologist at MD Anderson Cancer Center in Houston, USA
(b) a psychologist at RD Moursan Cancer Center in Hiroshima, Japan
(c) eye specialist at MD Anderson Eye Center in Berlin, Germany
(d) an anthropologist at AD Pitson Anthro Center in Tokyo, Japan
Answer:
(a) oncologist at MD Anderson Cancer Center in Houston, USA

Question 3.
Dr. Ritsuko is a radiation oncologist and also
(a) an advocate of civil cases
(b) an advocate of proton therapy
(c) an advocate of neuron therapy
(d) a judge at Hiroshima
Answer:
(b) an advocate of proton therapy

Question 4.
During the final stages of World War II in 1945, two atomic bombs were dropped on the cites of
(a) Hiroshima and Houston in Japan and USA
(b) Stalin and Berlin in Germany
(c) Hiroshima and Nagasaki in Japan
(d) None of the above
Answer:
(c) Hiroshima and Nagasaki in Japan

Question 5.
Name the first atomic bomb which was dropped in the city of Hiroshima and when?
(a) The Old Man, 26 August 1946
(b) The Young Boy 16 August 1954
(c) The Little Boy, 6 August 1945
(d) The Little Girl, 6 August 1944
Answer:
(c) The Little Boy, 6 August 1945

Question 6.
Name the second atomic bomb, which was dropped in the city of Nagasaki, and when?
(a) The Fat Woman, 9th Aug. 1945
(b) The Super Man, 9th Aug. 1954
(c) The Thin Man, 19th Sept. 1948
(d) The Fat Man, 9th Aug. 1945
Answer:
(d) The Fat Man, 9th Aug. 1945

Question 7.
Dr. Ritsuko Komaki was living with her family near :
(a) Sobosan
(b) Osaka
(c) Okayama
(d) Yamaguchi
Answer:
(b) Osaka

Question 8.
What vowed Komaki to become a cancer doctor?
(a) when her close friend Sadako Sasaki died at age 11 of radiation-related leukemia
(b) she developed both a fascination with and fear of radiation
(c) a long term effects which likely contributed heavily to the deaths of about half her relatives including her father
(d) both, (a) and (c)
Answer:
(a) when her close friend Sadako Sasaki died at age 11 of radiation-related leukemia

Question 9.
After becoming a cancer doctor, she uses it in increasingly sophisticated ways to :
(a) fight cancer
(b) fight diabetics
(c) fight leukemia
(d) fight bone marrow
Answer:
(a) fight cancer

Question 10.
Being the chief of the Clinical Section, she was also :
(a) the Program Director of Leukemia Radiation Oncology
(b) the Program Director of Thoracic Radiation Oncology
(c) Program Director of Bone-marrow
(d) none of the above
Answer:
(b) the Program Director of Thoracic Radiation Oncology

Question 11.
What does ‘leukemia’ mean?
(a) a disease
(b) throat cancer
(c) a type of cancer of the blood or bone marrow
(d) brain tumor
Answer:
(c) a type of cancer of the blood or bone marrow

Question 12.
Who do you think can we say as Oncologist?
(a) a doctor who treats bone marrow
(b) a doctor who treats rheumatism
(c) a doctor who treats cancer.
(d) none of the above
Answer:
(c) a doctor who treats cancer.

Question 13.
One who introduces new ideas of things
(a) inheritor
(b) intrigued
(c) inhibitor
(d) innovator
Answer:
(d) innovator

Question 14.
‘Compliance’ means
(a) obedience
(b) advanced
(c) obligation
(d) oblige
Answer:
(a) obedience

Question 15.
For which Komaki was awarded the Juan A. del Regato Foundation Gold Medal?
(a) for best radiation oncologist
(b) for best educator and teacher
(c) for the best expert in the treatment of cancer
(d) for the lady of the year
Answer:
(b) for best educator and teacher

Question 16.
Why does she try to educate younger people – students and trainees as well as patients?
(a) for the easy and early treatment of disease
(b) to make them know how to prevent diseases in life
(c) to live healthy lives and achieve their goal to help others
(d) none of the above
Answer:
(c) to live healthy lives and achieve their goal to help others

Unit – II
The text
Proton therapy …………… photon treatment.

Question 17.
What is proton therapy?
(a) Proton therapy, like other forms of radiation, aims ionizing particles onto a target tumor to damage and ultimately destroy its cancerous cells.
(b) Proton therapy, same as other forms of radiation, aims separated particles onto a target leukemia to destroy and ultimately damage its poisonous cells.
(c) both (a) and (b)
(d) none of the above
Answer:
(a) Proton therapy, like other forms of radiation, aims ionizing particles onto a target tumor to damage and ultimately destroy its cancerous cells.

Question 18.
___________are much more localized and powerful than the X-rays used in more established radiation therapies.
(a) neutron beams
(b) nuclear beams
(c) proton beams
(d) all of the above
Answer:
(c) proton beams

Question 19.
___________is big enough to kill the cancerous cells but some radiation also hits the healthy cells around the tumor,
(a) dose
(b) dosage
(c) drugs
(d) dorsal
Answer:
(b) dosage

Question 20.
Side effects of radiation are :
(a) pneumonitis
(b) esophagitis or lead to the growth of secondary cancers
(c) bone-marrow toxicity
(d) all of the above
Answer:
(d) all of the above

Question 21.
To reduce damage to healthy tissues by a scattered dose of low radiation, the oncologist requires :
(a) sharply delineated radiation
(b) proton beams, shaped to almost perfectly match the specific tumor
(c) aimed to strike it precisely
(d) all of the above
Answer:
(d) all of the above

Question 22.
Name the unit of electromagnetic energy which release up to 90 percent of the cancer-fighting energy, as they penetrate the skin.
(a) photon (X-rays)
(b) neutrons
(c) protons
(d) phototropism
Answer:
(a) photon (X-rays)

Question 23.
The success rate against prostate cancer, for example, is around :
(a) 90 to 95 percent
(b) 80 to 90 percent
(c) 95 to 100 percent
(d) 70 to 85 percent
Answer:
(a) 90 to 95 percent

Question 24.
Identify the name and occupation of Komaki’s husband.
(a) Dr. J. H. Frost, Head of the Division of Radiation Psychology at AT.M.D. Anderson Brain Academy
(b) Dr. John Marshall, Head of the division of Rheumatism at UTM.D Anderson Bone Center
(c) Dr. James Cox, Head of the division of Radiation Oncology at UT.M.D Anderson Cancer Center
(d) none of the above
Answer:
(c) Dr. James Cox, Head of the division of Radiation Oncology at UT.M.D Anderson Cancer Center

Question 25.
What do you mean by ‘efficacy’?
(a) efficiency
(b) effectiveness
(c) effort
(d) eradicate
Answer:
(b) effectiveness

Question 26.
‘Calibrated’ means :
(a) added
(b) calculated
(c) sum
(d) measured
Answer:
(d) measured

Question 27.
What does ‘collateral damage’ mean?
(a) injury to other organs
(b) injury to skeletal organs
(c) damage to all parts of the body
(d) all of the above
Answer:
(a) injury to other organs

Question 28.
The quality of being poisonous :
(a) toxicity
(b) tumor
(c) drugs
(d) none of the above
Answer:
(a) toxicity

Question 29.
Radiation therapy means :
(a) treatment of cancer by using radiation
(b) treatment of brain tumors by using radiation
(c) treatment of leukemia by using radiation
(d) treatment of rheumatism by using medicines and oils for external use only
Answer:
(a) treatment of cancer by using radiation

Unit – III
Warm-up
The Text
Surgery radiation …………. forefront.

Question 30.
Before proton therapy was adopted what were the methods used for treating cancer?
(a) surgery, radiation
(b) chemotherapy
(c) photon therapy
(d) all of the above
Answer:

Question 31.
Name the place where Komaki did her externship, internship, residency and fellowship and learned ‘how radiation could cure people’.
(a) Milwaukee
(b) Hiroshima
(c) Houston
(d) Nilwaukee
Answer:
(a) Milwauke

Question 32.
When did Komaki and her husband, Cox went to Columbia Presbyterian Medical Centre in Newyork and to work with whom ?
(a) 1986, Prof. Eric Hall
(b) 1985, Dr. Eric Hall
(c) 1998, Dr. Aric Hall
(d) 1958, Dr. Muric Mall
Answer:
(b) 1985, Dr. Eric Hall

Question 33.
Komaki began putting her years of research to work at UT.M. D. Anderson is
(a) 1989
(b) 1999
(c) 1988
(d) 1888
Answer:
(c) 1988

Question 34.
When did protons were discovered and by whom?
(a) protons were discovered by Ernest Rutherford in 1919
(b) protons were deciphered by Ymest Rutherford in 1981
(c) protons were discovered by Ernest Rutherford in 1920
(d) protons were discovered by Komaki in 1919
Answer:
(a) protons were discovered by Ernest Rutherford in 1919

Question 35.
Proton therapy didn’t commence until _____________ at Berkeley nuclear physics labs.
(a) 1945
(b) 1964
(c) 1953
(d) 1954
Answer:
(d) 1954

Question 36.
_____________ Laboratory partnered with Massachusetts General Hospital to begin treating cancer patients in 1961.
(a) The Harvard Cyclotron
(b) The Harvord Kyclotron
(c) The Warvord Cyclotron
(d) none of these
Answer:
(a) The Harvard Cyclotron

Question 37.
Proton Treatment Center opened at _____________ University Medical Center in southern California to offer the first hospital-based program.
(a) Loma Linda
(b) Oxford
(c) Cambridge
(d) Austria
Answer:
(a) Loma Linda

Question 38.
Whose efforts paid off in 2006, when the $ 125 million Proton Therapy Center opened its doors offering a complete range of proton treatments?
(a) Komaki and Dr. Eric Hall
(b) Komaki and Dr. Andrew Lee
(c) Komaki and her husband Cox
(d) none of the above
Answer:
(c) Komaki and her husband Cox

Question 39.
What is that therapy that requires a team of seven doctors, several physicians and dosimetrists, and a couple of dozen technicians in 10-15 patients daily with four of them taking proton therapy?
(a) Thoracic malignancies
(b) Oracic malignancies
(c) Thoranic maliganancious
(d) none of the above
Answer:
(a) Thoracic malignancies

Question 40.
Like other radiation treatments, proton therapy runs about :
(a) seven months
(b) seven hours
(c) seven days
(d) seven weeks
Answer:
(d) seven weeks

Question 41.
In all, the Proton Therapy Center, with Dr. Andrew Lee as medical director, treats :
(a) 76 to 86 patients daily
(b) 75 to 80 patients every month
(c) 75 to 80 patients every week
(d) 75 to 80 patients daily
Answer:
(d) 75 to 80 patients daily

Question 42.
What does secondary malignancy mean?
(a) cancer that arises in the background of another malignancy
(b) cancer that arises in the spot of another malignancy
(c) both (a) and (b)
(d) none of the above
Answer:
(a) cancer that arises in the background of another malignancy

Question 43.
Expert in dosage of radiation means:
(a) medicinal dose
(b) malignancy
(c) toxicity
(d) dosimetrist
Answer:
(d) dosimetris

Question 44.
Word ‘intrigued’ refers to :
(a) fearless
(b) very complicated
(c) interested to know more
(d) forming part of the basic nature of something
Answer:
(c) interested to know more

Question 45.
What do you understand by the term ‘commence’?
(a) recover
(b) begin
(c) conclusion
(d) restart
Answer:
(b) begin

Introducing the Author:
Dr. Ritsuko Komaki, a radiation cancer specialist at MD Anderson Cancer Centre in Houston, U.S.A., is an ardent champion of proton therapy.

About the Topic:
This article throws light on Dr. Ritsuko’s immense contribution towards the treatment of radiation-related ailments. She has a high opinion on proton therapy.

Summary:
The atomic bomb explosion on her native Hiroshima in 1945 had a profound effect on Ritsuko. The city was completely destroyed. She grew up to be a spectator of its devastating effect. He saw one tragedy after another: the deaths of about half her relatives, her father, and last of all, the death of her very intimate friend Sadako Sasaki at the age of 11 of radiation-related blood cancer. She could not remain silent. These three traumatic events made her take an oath to become a cancer doctor.

Now Ritsuko has learnt the method of applying, radiation in a creative manner. As a result, her fear of it becomes a thing of the past. Today Dr. Komaki, as clinical section chief and Program Director of Thoracic Radiation Oncology at The University of Texas M.D. Anderson Cancer Center in Houston is one of the world’s distinguished researchers and advocates of proton radiation beam therapy with which radiation-related cancer patients are treated.

Private hospitals are not suitable for the treatment of these patients on trial, because they lack adequate manpower, qualitative review boards, and maintenance of strict quality control of treatment. Komaki’s motto is to educate students, trainees, and patients to focus on living healthy lives. Besides, she wants to inspire them to reach their goal to help others. Ritsuko throws light on proton therapy. It is a form of treatment that damages and eventually destroys cancerous cells of the exact spot of the affected tumor with the help of charging particles.

There is a difference between proton therapy and photon therapy; the latter is not as good as the former. Proton therapy is confined to a particular area. It is powerful too. Photon therapy, in spite of killing cancerous cells, has side effects. It leads to an increase in secondary cancer. Proton therapy plays a great role in striking the specific tumor precisely. It results in the minimization of injury to other organs. Besides, photons (X-rays) emit 90 percent of their cancer-fighting energy, as they pierce the skin. But their utility is lost by 40 percent.

Proton therapy is more effective for patients who suffer from prostate cancer. Proton therapy is most useful for those whose localized cancer has not affected the distant parts of the body. It has a great effect on children in particular. Proton therapy also helps the elderly, because their excessively weak skin cannot resist the power of radiation therapy. Komaki’s patients think that proton therapy is fabulous because they can tolerate it more easily than photon treatment.

When Komai became a student of a medical school in Hiroshima, she learned that only surgery was a possible cancer cure. In the course of her externship, internship, residency, and fellowship in Milwaukee, Komaki came to know the importance of radiation in the treatment of cancer. Besides, she learnt that localized radiation therapy was not so harmful as chemotherapy. In 1985, she along with Cox set out for New York to work with Dr. Eric Hall, who was, at that time, a renowned global authority on the effects of the atomic bomb on human beings.

In 1988, she spent years of research work in this field at UT M.D. Anderson. Proton therapy started in 1954. The Harvard Cyclotron Laboratory in partnership with Massachusetts General Hospital started treating cancer patients. The necessary technology was very costly. Therefore, the treatment did not go beyond physics research labs until 1990. The first hospital-based program commenced with the opening of the Proton Treatment Center at Loma Linda University Medical Center in southern California.

Their keen interest to know more about proton therapy and their ceaseless efforts were eventually rewarded. The Proton Therapy Center came into existence in 2006, paving the way for providing a complete range of proton treatments. Today Komaki and her colleagues treat their patients with this therapy that requires many specialists. To Komaki, this is not the end of the road. She looks beyond the present. With a heart full of delight, she declares that this is the only place in the world having a scanning beam.

In her view, proton therapy is of great use to the future of cancer care. Komaki and her colleagues are trying to make the lives of the patients more comfortable than ever before, by destroying cancer cells, but not normal cells. The time has come when they lead a normal life during treatments. Dr. Ritsuko and her colleagues at UT M.D. Anderson may take a leading part in whatever takes place in this domain in future.

ସାରାଂଶ:
ଜାପାନର ହିରୋସୀମାରେ ୧୯୪୫ ମସିହାରେ ପଡ଼ିଥିବା ପରମାଣୁ ବୋମାର ପ୍ରଭାବ ରିକୋଙ୍କ ଉପରେ ଗଭୀର ଭାବେ ପ୍ରଭାବ ପକାଇଥିଲା । ପରମାଣୁ ବୋମା ସହରଟିକୁ ସମ୍ପୂର୍ଣ୍ଣଭାବେ ଧ୍ବସ୍ତବିଧ୍ବସ୍ତ କରିଦେଇଥିଲା । ଡାଃ ରିକୋ କୋମାକି ଏହି ଧ୍ୱଂସକାରୀ ପ୍ରଭାବର ପ୍ରତ୍ୟକ୍ଷଦର୍ଶୀରୂପେ ଧୀରେ ଧୀରେ ବଡ଼ ହୋଇଥିଲେ । ସେ ଗୋଟିଏ ଦୁଃଖଦ ଘଟଣା ପରେ ଅନ୍ୟ ଏକ ଦୁଃଖ ଘଟଣା ଦେଖୁଥିଲେ । ଅର୍ବାଧ‌ିକ ନିଜ ସମ୍ପର୍କୀୟମାନଙ୍କ ମୃତ୍ୟୁ, ନିଜ ବାପାଙ୍କ ମୃତ୍ୟୁ ଏବଂ ସର୍ବୋପରି ନିଜର ଅତି ଘନିଷ୍ଠ ବନ୍ଧୁ ସଡ଼ାକୋ ସାସାକିଙ୍କ ୧୧ ବର୍ଷ ବୟସରେ ଏହି ପରମାଣୁ ବୋମାର ତେଜସ୍କ୍ରିୟ ରଶ୍ମିର ପ୍ରଭାବରେ ରକ୍ତ କର୍କଟ ରୋଗ ଯୋଗୁଁ ମୃତ୍ୟୁ ତାଙ୍କୁ ବିବ୍ରତ ଓ ଅତିଷ୍ଠ କରିଦେଇଥିଲା । ସେ ନୀରବହୋଇ ରହିପାରିଲେ ନାହିଁ ।

ଏହି ତିନୋଟି ଦୁଃଖଦ ଘଟଣା ତାଙ୍କୁ ଜଣେ କର୍କଟ ରୋଗ ଚିକିତ୍ସକ ହେବାପାଇଁ ପ୍ରତିଜ୍ଞାବଦ୍ଧ କରାଇଥିଲା । ବର୍ତ୍ତମାନ ରିକୋ କୋମାକି ଏହି ତେଜସ୍କ୍ରିୟ ରଶ୍ମିର ସର୍ଜନାତ୍ମକ ପ୍ରୟୋଗ କରି ଚିକିତ୍ସା କରିପାରୁଛନ୍ତି । ଫଳରେ ତାଙ୍କର ଭୟ ଦୂର ହୋଇପାରିଛି । ବର୍ତ୍ତମାନ ସେ ହୋଷ୍ଟନରେ ଥିବା Taxas M.D. Anderson Cancer Centreର Thoracic Radiation Oncologyର ମୁଖ୍ୟ ଓ ପ୍ରୋଗ୍ରାମ ନିର୍ଦ୍ଦେଶକ ଭାବେ ବିଶ୍ଵର ସ୍ୱନାମଧନ୍ୟ ଗବେଷକ ଏବଂ ପ୍ରୋଟନ ରଶ୍ମିର ପ୍ରୟୋଗରେ କର୍କଟ ବ୍ୟାଧ୍ ଚିକିତ୍ସା ପଦ୍ଧତିର ସମର୍ଥକମାନଙ୍କ ମଧ୍ୟରେ ଜଣେ ଭାବେ ଗଣା ହେଉଛନ୍ତି । ତାଙ୍କ ମତରେ ଘରୋଇ ଚିକିତ୍ସାଳୟଗୁଡ଼ିକ ଚିକିତ୍ସା ଓ ପରୀକ୍ଷା ପାଇଁ ଅନୁପଯୁକ୍ତ, କାରଣ ସେମାନଙ୍କ ପାଖରେ ଆବଶ୍ୟକୀୟ ଦକ୍ଷ ଚିକିତ୍ସକ ଗୁଣାତ୍ମକ ସମୀକ୍ଷାମଣ୍ଡଳୀ, ଏବଂ ଗୁଣାତ୍ମକ ଚିକିତ୍ସା ପଦ୍ଧତିର ଅଭାବ ରହିଛି ।

ଶିକ୍ଷାର୍ଥୀ, ତାଲିମ ପାଉଥ‌ିବା ବ୍ୟକ୍ତି ଏବଂ ରୋଗୀମାନଙ୍କୁ ସ୍ବାସ୍ଥ୍ୟକର ଜୀବନ ବଞ୍ଚିବା ପାଇଁ ଶିକ୍ଷା ଦେବାକୁ ସେମାନଙ୍କ ଲକ୍ଷ୍ୟ ରହିଛି । ଏହାଛଡ଼ା କୋମାକି ଚାହାନ୍ତି ସେମାନଙ୍କ ଲକ୍ଷ୍ୟପ୍ରାପ୍ତି ପାଇଁ ସେମାନେ ଅନ୍ୟମାନଙ୍କୁ ସାହାଯ୍ୟ କରିବା ଉପରେ ଗୁରୁତ୍ଵ ଦେବା ଉଚିତ । ରିଟ୍ସ୍କୋ ପ୍ରୋଟନ୍ ଚିକିତ୍ସା ପଦ୍ଧତି ଉପରେ ଆଲୋକପାତ କରିଛନ୍ତି । ଏହି ଚିକିତ୍ସା ପଦ୍ଧତି ମାଧ୍ୟମରେ ଟ୍ୟୁମରର ନିର୍ଦ୍ଦିଷ୍ଟ କ୍ଷତିଗ୍ରସ୍ତ ସ୍ଥାନରେ କର୍କଟ ସଂକ୍ରମିତ କୋଷଗୁଡ଼ିକୁ ଚାର୍ଜିତ କଣିକାଗୁଡ଼ିକ ସାହାଯ୍ୟରେ ନଷ୍ଟ କରି ଦିଆଯାଏ । ଫୋଟନ୍ ଚିକିତ୍ସା ପ୍ରୋଟନ୍‌ ଚିକିତ୍ସା ଅପେକ୍ଷା ଉତ୍ତମ ନୁହେଁ । ପ୍ରୋଟନ୍ ଚିକିତ୍ସା ଏକ ନିର୍ଦ୍ଦିଷ୍ଟ ସ୍ଥାନରେ ସୀମାବଦ୍ଧ ରହେ । ଏହା ମଧ୍ୟ ଅଧ‌ିକ ଶକ୍ତିଶାଳୀ । ଫୋଟନ୍ ଚିକିତ୍ସା ପଦ୍ଧତି କର୍କଟ କୋଷଗୁଡ଼ିକୁ ନଷ୍ଟ କରୁଥିବା ସତ୍ତ୍ବେ ଏହାର ବିଭିନ୍ନ ପାର୍ଶ୍ଵ ପ୍ରତିକ୍ରିୟା ଥାଏ ।

ଏହା କର୍କଟ ରୋଗକୁ ବଢ଼େଇ ଦେବାରେ ସହାୟକ ହୋଇଥାଏ । ପ୍ରୋଟନ୍ ଚିକିତ୍ସା ମାଧ୍ୟମରେ ଟ୍ୟୁମରକୁ ନିର୍ଦ୍ଦିଷ୍ଟଭାବେ ଆଘାତ କରାଯାଇପାରେ । ଅନ୍ୟ ଅଙ୍ଗପ୍ରତ୍ୟଙ୍ଗଗୁଡ଼ିକ ଅପାତତଃ କମ୍ କ୍ଷତି ହୋଇଥାଏ । ଫୋଟନ ଚିକିତ୍ସା ପଦ୍ଧତିରେ ବିକିରିତ ରଶ୍ମି ଶରୀରରେ ପ୍ରବେଶ କରିବା ସମୟରେ ୨୦ ପ୍ରତିଶତ କର୍କଟ ରୋଗ ପ୍ରତିରୋଧକ ଶକ୍ତି ଉତ୍ପନ୍ନ କରିଥାଏ । କିନ୍ତୁ ତା’ର 40 ପ୍ରତିଶତ ଉପଯୋଗିତା ନଷ୍ଟ ହୋଇଥଯାଏ । ପ୍ରାରମ୍ଭିକ ପର୍ଯ୍ୟାୟର କର୍କଟ ରୋଗୀମାନଙ୍କ ପାଇଁ ପ୍ରୋଟନ୍ ଚିକିତ୍ସା ପଦ୍ଧତି ଅଧ‌ିକ ଫଳପ୍ରସୂ । ଯେଉଁମାନଙ୍କର ନିର୍ଦ୍ଦିଷ୍ଟ ଅଂଶ କର୍କଟ ସଂକ୍ରମିତ ମାତ୍ର ଅନ୍ୟ ଅଂଶ ସୁସ୍ଥ ଅଛି ସେମାନଙ୍କ ପାଇଁ ଏହି ପଦ୍ଧତି ଅଧ୍ଵ ଫଳପ୍ରଦ । ପିଲାମାନଙ୍କ ପାଇଁ ଏହା ଅଧିକ ଫଳପ୍ରଦ ହୋଇଥାଏ ।

ବୟସ୍କମାନଙ୍କ ପାଇଁ ମଧ୍ୟ ଏହା ଲାଭଦାୟକ କାରଣ ସେମାନଙ୍କ ଅତି ଦୁର୍ବଳ ଚର୍ମ ବିକିରଣ ପଦ୍ଧତିର ତୀବ୍ରତାକୁ ସହ୍ୟ କରିପାରେ ନାହିଁ । କୋମାକିଙ୍କର ରୋଗୀମାନେ ଭାବନ୍ତି ଯେ ପ୍ରୋଟନ୍ ଚିକିତ୍ସା ହେଉଛି ସୁଖକର କାରଣ ସେମାନେ ଏହାକୁ ଫୋଟନ୍ ଚିକିତ୍ସା ଅପେକ୍ଷା ସହଜରେ ସହ୍ୟ କରିପାରନ୍ତି । ଯେତେବେଳେ କୋମାକି ହିରୋସୀମା ମେଡ଼ିକାଲ ବିଦ୍ୟାଳୟର ଛାତ୍ରୀ ଥିଲେ, କର୍କଟ ରୋଗର କେବଳ ଅସ୍ତ୍ରୋପଚାର ମାଧ୍ୟମରେ ସମ୍ଭବ ବୋଲି ଶିକ୍ଷାଲାଭ କରିଥିଲେ । Milwaukeeଠାରେ ସେ କର୍କଟ ରୋଗର ଚିକିତ୍ସା କ୍ଷେତ୍ରରେ ବିକିରଣର ଗୁରୁତ୍ଵ ବିଷୟରେ ଜାଣିଲେ l ଏହା ବ୍ୟତୀତ ସେ ଜାଣିଲେ ଯେ ବିକିରଣ ଚିକିତ୍ସା ଠାରୁ କମ୍ କ୍ଷତିକାରକ ଅଟେ । ୧୯୮୫ ମସିହାରେ ସେ Coxଙ୍କ ସହିତ New York ଅଭିମୁଖେ ଯାତ୍ରା କଲେ Dr. Eric Hallଙ୍କ ସହିତ କାମ କରିବା ପାଇଁ ଯେ କି ମନୁଷ୍ୟ ସମାଜ ଉପରେ ପରମାଣୁ ବୋମାର ପ୍ରଭାବ ଉପରେ କାର୍ଯ୍ୟ କରୁଥିଲେ । ୧୯୮୮ରେ UT M.D. Andersonଠାରେ ଏହି ସମ୍ବନ୍ଧରେ ଗବେଷଣା କରିଥିଲେ ।

ପ୍ରୋଟନ୍ ଚିକିତ୍ସା ଆରମ୍ଭ ହେଲା ୧୯୪୫ରେ । The Havard Cyclotron Laboratory ଓ Massachussetts General Hospital ଏକତ୍ର ମିଶି କର୍କଟ ରୋଗୀମାନଙ୍କ ଚିକିତ୍ସା ଆରମ୍ଭ କଲେ । ଏଥିରେ ବ୍ୟବହୃତ ଯନ୍ତ୍ରପାତିଗୁଡ଼ିକ ବହୁତ ବ୍ୟୟବହୁଳ ଥିଲା ! ତେଣୁ ଏହି ଚିକିତ୍ସା ପଦାର୍ଥ ବିଜ୍ଞାନର ଗବେଷଣାଗାର ବାହାରକୁ ୧୯୯୦ ମସିହା ପର୍ଯ୍ୟନ୍ତ ଯାଇପାରି ନ ଥିଲା । ଦକ୍ଷିଣ କାଲିଫଣ୍ଡିଆର Loma Linda University Medical Centreରେ ପ୍ରୋଟନ୍ ଚିକିତ୍ସା କେନ୍ଦ୍ର ଆରମ୍ଭ ହୋଇଥିଲା । ପରିଶେଷରେ ସେମାନଙ୍କର ପରିଶ୍ରମର ଉପଯୁକ୍ତ ପୁରସ୍କାର ସେମାନେ ପାଇଥିଲେ । ପ୍ରୋଟନ୍‌ ଚିକିତ୍ସା କେନ୍ଦ୍ର ୨୦୦୬ ମସିହାରେ ଆରମ୍ଭ ହୋଇଥିଲା ।

ଏବେ କୋମାକି ଏବଂ ତାଙ୍କର ସହକର୍ମୀମାନେ ଏହି ଚିକିତ୍ସାରେ ନିଯୁକ୍ତ ଅଛନ୍ତି । କୋମାକିଙ୍କ ମତରେ ଏହା ଆରମ୍ଭ ମାତ୍ର । ସେ ଭବିଷ୍ୟତ ଉପରେ ଦୃଷ୍ଟି ନିବଦ୍ଧ କରିଛନ୍ତି । ସେ ଗର୍ବର ସହ କହନ୍ତି ଯେ କର୍କଟ ରୋଗର ସଫଳ ବିକିରଣ ଚିକିତ୍ସା ପାଇଁ ତାହା ହିଁ ଏକମାତ୍ର ଚିକିତ୍ସାଳୟ । ତାଙ୍କ ମତରେ କର୍କଟ ବ୍ୟାଧର ଆରୋଗ୍ୟ ନିମନ୍ତେ ପ୍ରୋଟନ ଚିକିତ୍ସା ପଦ୍ଧତି ଉପାଦେୟତା ଖୁବ୍ ବେଶି । କୋମାକି ଓ ତାଙ୍କ ସହକର୍ମୀମାନେ କର୍କଟରୋଗୀମାନଙ୍କର କେବଳ କ୍ୟାନସର ସଂକ୍ରମିତ କୋଷଗୁଡ଼ିକୁ ନଷ୍ଟ କରିଦେଇ ସେମାନଙ୍କୁ ଆରାମପ୍ରଦ ଜୀବନ ବଞ୍ଚିବାକୁ ସୁଯୋଗ ଦେବାପାଇଁ ଉଦ୍ୟମ କରୁଛନ୍ତି । ସମୟ ଆସିବ ଯେତେବେଳେ ସେମାନେ ଚିକିତ୍ସା ସମୟରେ ମଧ୍ୟ ସାଧାରଣ ଜୀବନଯାପନ କରିପାରିବେ । କର୍କଟ ବ୍ୟାଧ୍ ଚିକିତ୍ସା କ୍ଷେତ୍ରରେ ଭବିଷ୍ୟତରେ ଯାହାକି ଅଗ୍ରଗତି ହେବ ସେଥୁରେ କୋମାକି ଏବଂ ତାଙ୍କ ସହକର୍ମୀମାନେ ହିଁ ଅଗ୍ରଣୀ ଭୂମିକା ନେବାର ସମ୍ଭାବନା ଅଛି ।