RD Sharma Solutions for Class 11 Maths Exercise 24.2 Chapter 24 - The Circle

In Exercise 24.2 of Chapter 24, we shall discuss problems based on the concepts of the general equation of a circle. Students who wish to strengthen their knowledge and build a strong command over the subject can refer to RD Sharma Class 11 Maths Solutions. Experts tutors have designed the solutions in a simple and understandable language, in order to meet the requirements of students and help them score well in the board exams. Here, the solutions to this exercise are provided in pdf format, which can be downloaded easily from the links given below.

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Access answers to RD Sharma Solutions for Class 11 Maths Exercise 24.2 Chapter 24 – The Circle

1. Find the coordinates of the centre radius of each of the following circle:
(i) x² + y² + 6x – 8y – 24 = 0

(ii) 2x² + 2y² – 3x + 5y = 7

(iv) x² + y² – ax – by = 0

Solution:

(i) x² + y² + 6x – 8y – 24 = 0

Given:

The equation of the circle is x² + y² + 6x – 8y – 24 = 0 …… (1)

We know that for a circle x² + y² + 2ax + 2by + c = 0 …… (2)

Centre = (-a, -b)

So by comparing equation (1) and (2)

Centre = (-6/2, -(-8)/2)

= (-3, 4)

Radius = √(a² + b² – c)

= √(3² + 4² – (-24))

= √(9 + 16 + 24)

= √(49)

= 7

∴ The centre of the circle is (-3, 4) and the radius is 7.

(ii) 2x² + 2y² – 3x + 5y = 7

Given:

The equation of the circle is 2x² + 2y² – 3x + 5y = 7 (divide by 2 we get)

x² + y² – 3x/2 + 5y/2 = 7/2

Now, by comparing with the equation x² + y² + 2ax + 2by + c = 0 

Centre = (-a, -b)

Given:

The equation of the circle is

(Multiply by 2 we get)

x² + y² + 2x cos θ + 2y sin θ – 8 = 0

By comparing with the equation x² + y² + 2ax + 2by + c = 0

Centre = (-a, -b)

= [(-2cos θ)/2 , (-2sin θ)/2]

= (-cos θ, -sin θ)

Radius = √(a² + b² – c)

= √[(-cos θ)² + (sin θ)² – (-8)]

= √[cos² θ + sin² θ + 8]

= √[1 + 8]

= √[9]

= 3

∴ The centre and radius of the circle is (-cos θ, -sin θ) and 3.

(iv) x² + y² – ax – by = 0

Given:

Equation of the circle is x² + y² – ax – by = 0

By comparing with the equation x² + y² + 2ax + 2by + c = 0

Centre = (-a, -b)

= (-(-a)/2, -(-b)/2)

= (a/2, b/2)

Radius = √(a² + b² – c)

= √[(a/2)² + (b/2)²]

= √[(a²/4 + b²/4)]

= √[(a² + b²)/4]

= [√(a² + b²)]/2

∴ The centre and radius of the circle is (a/2, b/2) and [√(a² + b²)]/2

2. Find the equation of the circle passing through the points :
(i) (5, 7), (8, 1) and (1, 3)

(ii) (1, 2), (3, – 4) and (5, – 6)

(iii) (5, -8), (-2, 9) and (2, 1)

(iv) (0, 0), (-2, 1) and (-3, 2)

Solution:

(i) (5, 7), (8, 1) and (1, 3)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

Firstly let us find the values of a, b and c

Substitute the given point (5, 7) in equation (1), we get

5² + 7² + 2a (5) + 2b (7) + c = 0

25 + 49 + 10a + 14b + c = 0

10a + 14b + c + 74 = 0….. (2)

By substituting the given point (8, 1) in equation (1), we get

8² + 1² + 2a (8) + 2b (1) + c = 0

64 + 1 + 16a + 2b + c = 0

16a + 2b + c + 65 = 0….. (3)

Substituting the point (1, 3) in equation (1), we get

1² + 3² + 2a (1) + 2b (3) + c = 0

1 + 9 + 2a + 6b + c = 0

2a + 6b + c + 10 = 0….. (4)

Now by simplifying the equations (2), (3), (4) we get the values

a = -29/6, b = -19/6, c = 56/3

Substituting the values of a, b, c in equation (1), we get

x² + y² + 2 (-29/6)x + 2 (-19/6) + 56/3 = 0

x² + y² – 29x/3 – 19y/3 + 56/3 = 0

3x² + 3y² – 29x – 19y + 56 = 0

∴ The equation of the circle is 3x² + 3y² – 29x – 19y + 56 = 0

(ii) (1, 2), (3, – 4) and (5, – 6)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

Substitute the points (1, 2) in equation (1), we get

1² + 2² + 2a (1) + 2b (2) + c = 0

1 + 4 + 2a + 4b + c = 0

2a + 4b + c + 5 = 0….. (2)

Substitute the points (3, -4) in equation (1), we get

3² + (- 4)² + 2a (3) + 2b (- 4) + c = 0

9 + 16 + 6a – 8b + c = 0

6a – 8b + c + 25 = 0….. (3)

Substitute the points (5, -6) in equation (1), we get

5² + (- 6)² + 2a (5) + 2b (- 6) + c = 0

25 + 36 + 10a – 12b + c = 0

10a – 12b + c + 61 = 0….. (4)

Now by simplifying the equations (2), (3), (4) we get

a = – 11, b = – 2, c = 25

Substitute the values of a, b and c in equation (1), we get

x² + y² + 2(- 11)x + 2(- 2) + 25 = 0

x² + y² – 22x – 4y + 25 = 0

∴ The equation of the circle is x² + y² – 22x – 4y + 25 = 0

(iii) (5, -8), (-2, 9) and (2, 1)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

Substitute the point (5, -8) in equation (1), we get

5² + (- 8)² + 2a(5) + 2b(- 8) + c = 0

25 + 64 + 10a – 16b + c = 0

10a – 16b + c + 89 = 0….. (2)

Substitute the points (-2, 9) in equation (1), we get

(- 2)² + 9² + 2a(- 2) + 2b(9) + c = 0

4 + 81 – 4a + 18b + c = 0

-4a + 18b + c + 85 = 0….. (3)

Substitute the points (2, 1) in equation (1), we get

2² + 1² + 2a(2) + 2b(1) + c = 0

4 + 1 + 4a + 2b + c = 0

4a + 2b + c + 5 = 0….. (4)

By simplifying equations (2), (3), (4) we get

a = 58, b = 24, c = – 285.

Now, by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(58)x + 2(24) – 285 = 0

x² + y² + 116x + 48y – 285 = 0

∴ The equation of the circle is x² + y² + 116x + 48y – 285 = 0

(iv) (0, 0), (-2, 1) and (-3, 2)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (1)

Substitute the points (0, 0) in equation (1), we get

0² + 0² + 2a(0) + 2b(0) + c = 0

0 + 0 + 0a + 0b + c = 0

c = 0….. (2)

Substitute the points (-2, 1) in equation (1), we get

(- 2)² + 1² + 2a(- 2) + 2b(1) + c = 0

4 + 1 – 4a + 2b + c = 0

-4a + 2b + c + 5 = 0….. (3)

Substitute the points (-3, 2) in equation (1), we get

(- 3)² + 2² + 2a(- 3) + 2b(2) + c = 0

9 + 4 – 6a + 4b + c = 0

-6a + 4b + c + 13 = 0….. (4)

By simplifying the equations (2), (3), (4) we get

a = -3/2, b = -11/2, c = 0

Now, by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(-3/2)x + 2(-11/2)y + 0 = 0

x² + y² – 3x – 11y = 0

∴ The equation of the circle is x² + y² – 3x – 11y = 0

3. Find the equation of the circle which passes through (3, – 2), (- 2, 0) and has its centre on the line 2x – y = 3.

Solution:

Given:

The line 2x – y = 3 … (1)

The points (3, -2), (-2, 0)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (2)

Let us substitute the centre (-a, -b) in equation (1) we get,

2(- a) – (- b) = 3

-2a + b = 3

2a – b + 3 = 0…… (3)

Now Substitute the given points (3, -2) in equation (2), we get

3² + (- 2)² + 2a(3) + 2b(- 2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (4)

Substitute the points (-2, 0) in equation (2), we get

(- 2)² + 0² + 2a(- 2) + 2b(0) + c = 0

4 + 0 – 4a + c = 0

4a – c – 4 = 0….. (5)

By simplifying the equations (3), (4) and (5) we get,

a = 3/2, b = 6, c = 2

Again by substituting the values of a, b, c in (2), we get

x² + y² + 2 (3/2)x + 2 (6)y + 2 = 0

x² + y² + 3x + 12y + 2 = 0

∴ The equation of the circle is x² + y² + 3x + 12y + 2 = 0.

4. Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1.

Solution:

Given:

The points (3, 7), (5, 5)

The line x – 4y = 1…. (1)

By using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0 ….. (2)

Let us substitute the centre (-a, -b) in equation (1) we get,

(- a) – 4(- b) = 1

-a + 4b = 1

a – 4b + 1 = 0…… (3)

Substitute the points (3, 7) in equation (2), we get

3² + 7² + 2a(3) + 2b(7) + c = 0

9 + 49 + 6a + 14b + c = 0

6a + 14b + c + 58 = 0….. (4)

Substitute the points (5, 5) in equation (2), we get

5² + 5² + 2a(5) + 2b(5) + c = 0

25 + 25 + 10a + 10b + c = 0

10a + 10b + c + 50 = 0….. (5)

By simplifying equations (3), (4) and (5) we get,

a = 3, b = 1, c = – 90

Now, by substituting the values of a, b, c in equation (2), we get

x² + y² + 2(3)x + 2(1)y – 90 = 0

x² + y² + 6x + 2y – 90 = 0

∴ The equation of the circle is x² + y² + 6x + 2y – 90 = 0.

5. Show that the points (3, -2), (1, 0), (-1, -2) and (1, -4) are con – cyclic.

Solution:

Given:

The points (3, -2), (1, 0), (-1, -2) and (1, -4)

Let us assume the circle passes through the points A, B, C.

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points A (3, – 2) in equation (1), we get,

3² + (-2)² + 2a(3) + 2b(-2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (2)

Substitute the points B (1, 0) in equation (1), we get,

1² + 0² + 2a(1) + 2b(0) + c = 0

1 + 2a + c = 0 ……- (3)

Substitute the points C (-1, -2) in equation (1), we get,

(- 1)² + (- 2)² + 2a(- 1) + 2b(- 2) + c = 0

1 + 4 – 2a – 4b + c = 0

5 – 2a – 4b + c = 0

2a + 4b – c – 5 = 0….. (4)

Upon simplifying the equations (2), (3) and (4) we get,

a = – 1, b = 2 and c = 1

Substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 1)x + 2(2)y + 1 = 0

x² + y² – 2x + 4y + 1 = 0 ….. (5)

Now by substituting the point D (1, -4) in equation (5) we get,

1² + (- 4)² – 2(1) + 4(- 4) + 1

1 + 16 – 2 – 16 + 1

0

∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) are con – cyclic.

6. Show that the points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle, and find its equation, centre, and radius.

Solution:

Given:

The points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle.

Let us assume the circle passes through the points A, B, C.

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substituting A (5, 5) in (1), we get,

5² + 5² + 2a(5) + 2b(5) + c = 0

25 + 25 + 10a + 10b + c = 0

10a + 10b + c + 50 = 0….. (2)

Substitute the points B (6, 4) in equation (1), we get,

6² + 4² + 2a(6) + 2b(4) + c = 0

36 + 16 + 12a + 8b + c = 0

12a + 8b + c + 52 = 0….. (3)

Substitute the point C (-2, 4) in equation (1), we get,

(-2)² + 4² + 2a(-2) + 2b(4) + c = 0

4 + 16 – 4a + 8b + c = 0

20 – 4a + 8b + c = 0

4a – 8b – c – 20 = 0….. (4)

Upon simplifying equations (2), (3) and (4) we get,

a = – 2, b = – 1 and c = – 20

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 2)x + 2(- 1)y – 20 = 0

x² + y² – 4x – 2y – 20 = 0 ….. (5)

Substituting D (7, 1) in equation (5) we get,

7² + 1² – 4(7) – 2(1) – 20

49 + 1 – 28 – 2 – 20

0

∴ The points (3, -2), (1, 0), (-1, -2), (1, -4) lie on a circle.

Now let us find the centre and the radius.

We know that for a circle x² + y² + 2ax + 2by + c = 0,

Centre = (-a, -b)

Radius = √(a² + b² – c) 

Comparing equation (5) with equation (1), we get

Centre = [-(-4)/2, -(-2)/2)]

= (2, 1)

Radius = √(2² + 1² – (-20)) 

= √(25)

= 5

∴ The centre and radius of the circle is (2, 1) and 5.

7. Find the equation of the circle which circumscribes the triangle formed by the lines:
(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3

(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0

(iii) x + y = 2, 3x – 4y = 6 and x – y = 0

(iv) y = x + 2, 3y = 4x and 2y = 3x

Solution:

(i) x + y + 3 = 0, x – y + 1 = 0 and x = 3

Given:

The lines x + y + 3 = 0

x – y + 1 = 0

x = 3

On solving these lines we get the intersection points A (-2, -1), B (3, 4), C (3, -6)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (-2, -1) in equation (1), we get

(- 2)² + (- 1)² + 2a(-2) + 2b(-1) + c = 0

4 + 1 – 4a – 2b + c = 0

5 – 4a – 2b + c = 0

4a + 2b – c – 5 = 0….. (2)

Substitute the points (3, 4) in equation (1), we get

3² + 4² + 2a(3) + 2b(4) + c = 0

9 + 16 + 6a + 8b + c = 0

6a + 8b + c + 25 = 0….. (3)

Substitute the points (3, -6) in equation (1), we get

3² + (- 6)² + 2a(3) + 2b(- 6) + c = 0

9 + 36 + 6a – 12b + c = 0

6a – 12b + c + 45 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = – 3, b = 1, c = -15.

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 3)x + 2(1)y – 15 = 0

x² + y² – 6x + 2y – 15 = 0

∴ The equation of the circle is x² + y² – 6x + 2y – 15 = 0.

(ii) 2x + y – 3 = 0, x + y – 1 = 0 and 3x + 2y – 5 = 0

Given:

The lines 2x + y – 3 = 0

x + y – 1 = 0

3x + 2y – 5 = 0

On solving these lines we get the intersection points A(2, – 1), B(3, – 2), C(1,1)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (2, -1) in equation (1), we get

2² + (- 1)² + 2a(2) + 2b(- 1) + c = 0

4 + 1 + 4a – 2b + c = 0

4a – 2b + c + 5 = 0….. (2)

Substitute the points (3, -2) in equation (1), we get

3² + (- 2)² + 2a(3) + 2b(- 2) + c = 0

9 + 4 + 6a – 4b + c = 0

6a – 4b + c + 13 = 0….. (3)

Substitute the points (1, 1) in equation (1), we get

1² + 1² + 2a(1) + 2b(1) + c = 0

1 + 1 + 2a + 2b + c = 0

2a + 2b + c + 2 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = -13/2, b = -5/2, c = 16

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2 (-13/2)x + 2 (-5/2)y + 16 = 0

x² + y² – 13x – 5y + 16 = 0

∴ The equation of the circle is x² + y² – 13x – 5y + 16 = 0

(iii) x + y = 2, 3x – 4y = 6 and x – y = 0

Given:

The lines x + y = 2

3x – 4y = 6

x – y = 0

On solving these lines we get the intersection points A(2,0), B(- 6, – 6), C(1,1)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (2, 0) in equation (1), we get

2² + 0² + 2a(2) + 2b(0) + c = 0

4 + 4a + c = 0

4a + c + 4 = 0….. (2)

Substitute the point (-6, -6) in equation (1), we get

(- 6)² + (- 6)² + 2a(- 6) + 2b(- 6) + c = 0

36 + 36 – 12a – 12b + c = 0

12a + 12b – c – 72 = 0….. (3)

Substitute the points (1, 1) in equation (1), we get

1² + 1² + 2a(1) + 2b(1) + c = 0

1 + 1 + 2a + 2b + c = 0

2a + 2b + c + 2 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = 2, b = 3, c = – 12.

Substituting the values of a, b, c in equation (1), we get

x² + y² + 2(2)x + 2(3)y – 12 = 0

x² + y² + 4x + 6y – 12 = 0

∴ The equation of the circle is x² + y² + 4x + 6y – 12 = 0

(iv) y = x + 2, 3y = 4x and 2y = 3x

Given:

The lines y = x + 2

3y = 4x

2y = 3x

On solving these lines we get the intersection points A(6,8), B(0,0), C(4,6)

So by using the standard form of the equation of the circle:

x² + y² + 2ax + 2by + c = 0….. (1)

Substitute the points (6, 8) in equation (1), we get

6² + 8² + 2a(6) + 2b(8) + c = 0

36 + 64 + 12a + 16b + c = 0

12a + 16b + c + 100 = 0…… (2)

Substitute the points (0, 0) in equation (1), we get

0² + 0² + 2a(0) + 2b(0) + c = 0

0 + 0 + 0a + 0b + c = 0

c = 0….. (3)

Substitute the points (4, 6) in equation (1), we get

4² + 6² + 2a(4) + 2b(6) + c = 0

16 + 36 + 8a + 12b + c = 0

8a + 12b + c + 52 = 0….. (4)

Upon simplifying equations (2), (3), (4) we get

a = – 23, b = 11, c = 0

Now by substituting the values of a, b, c in equation (1), we get

x² + y² + 2(- 23)x + 2(11)y + 0 = 0

x² + y² – 46x + 22y = 0

∴ The equation of the circle is x² + y² – 46x + 22y = 0