Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B System of Circles Important Questions Short Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2B System of Circles Important Questions Short Answer Type

Question 1.

Find the equation of the circle which passes through the origin and intersects the circles x^{2} + y^{2} – 4x – 6y – 3 = 0, x^{2} + y^{2} – 8y + 12 = 0 orthogonally.

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 …………(1)

Since (1) passes through the point (0, 0), then

(0)^{2} + (0)^{2} + 2g(0) + 2f(0) + c = 0

⇒ c = 0

Given equations of the circles are

x^{2} + y^{2} – 4x – 6y – 3 = 0 ……..(2)

x^{2} + y^{2} – 8y + 12 = 0 ……….(3)

Since (1) and (2) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(-2) + 2f(-3) = 0 – 3

⇒ -4g – 6f = -3

⇒ 4g + 6f – 3 = 0 …….(4)

Since (1) and (3) are orthogonal, then

2gg’ + 2ff’ = c + c’

⇒ 2g(0) + 2f(-4) = 0 + 12

⇒ -8f = 12

⇒ f = \(\frac{-3}{2}\)

Substituting the value of ‘f’ in (4)

4g + 6(\(\frac{-3}{2}\)) – 3 = 0

⇒ 4g – 9 – 3 = 0

⇒ 4g = 12

⇒ g = 3

∴ The equation of the required circle is

x^{2} + y^{2} + 2(3)x + 2(\(\frac{-3}{2}\))y + 0 = 0

x^{2} + y^{2} + 6x – 3y = 0

Question 2.

Find the equation of the circle which passes through the origin and intersects the circles x^{2} + y^{2} – 4x + 6y + 10 = 0, x^{2} + y^{2} + 12y + 6 = 0 orthogonally. [(AP) May ’18]

Solution:

Let, the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……..(1)

Since, (1) passes through the point (0, 0) then

(0)^{2} + (0)^{2} + 2g(0) + 2f(0) + c = 0

⇒ c = 0

Given equations of the circles are

x^{2} + y^{2} – 4x + 6y + 10 = 0 ……..(2)

x^{2} + y^{2} + 12y + 6 = 0 …….(3)

Since (1) and (2) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2(g)(-2) + 2f(3) = c + 10

⇒ -4g + 6f = 0 + 10

⇒ -4g + 6f = 10

⇒ 2g – 3f = -5

⇒ 2g – 3f + 5 = 0 ……..(4)

Since (1) and (3) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(0) + 2f(6) = 0 + 6

⇒ 12f = 6

⇒ f = \(\frac{1}{2}\)

Substituting the value of f in (4)

2g – 3(\(\frac{1}{2}\)) + 5 = 0

⇒ 2g + \(\frac{7}{2}\) = 0

⇒ g = \(\frac{-7}{4}\)

∴ The equation of the required circle is from (1)

x^{2} + y^{2} + 2(\(\frac{-7}{4}\))x + 2(\(\frac{1}{2}\))y + 0 = 0

⇒ x^{2} + y^{2} – \(\frac{7}{2}\)x + y = 0

⇒ 2x^{2} + 2y^{2} – 7x + 2y = 0

Question 3.

Find the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles x^{2} + y^{2} – 8x – 2y + 16 = 0 and x^{2} + y^{2} – 4x – 4y – 1 = 0.

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(1)

Since (1) passes through point (1, 1) then

(1)^{2} + (1)^{2} – 2g(1) – 2f(1) + c = 0

⇒ 2 + 2g + 2f + c = 0

⇒ 2g + 2f + c = -2 …….(2)

Given equations of the circles are

x^{2} + y^{2} – 8x – 2y + 16 = 0 ……..(3)

x^{2} + y^{2} – 4x – 4y – 1 = 0 ………(4)

Since (1) and (3) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(-4) + 2f(-1) = c + 16

⇒ -8g – 2f – c = 16 ………(5)

Since (1) and (4) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(-2) + 2f(-2) = c – 1

⇒ -4g – 4f – c = -1

From (2) and (5)

2g + 2f + c = -2

⇒ -8g – 2f – c = 16

⇒ -6g = 14

⇒ g = \(\frac{-7}{3}\)

From (5) and (6)

Substituting the values of g, f in (2)

∴ The equation of the required circle is from (1),

x^{2} + y^{2} + 2(\(\frac{-7}{3}\))x + 2(\(\frac{23}{6}\))y – 5 = 0

⇒ 3x^{2} + 3y^{2} – 14x + 23y – 15 = 0

Question 4.

Find the equation of the circle which passes through the point (0, -3) and intersects the circles given by the equations x^{2} + y^{2} – 6x + 3y + 5 = 0 and x^{2} + y^{2} – x – 7y = 0 orthogonally. [May ’15 (TS) May ’13]

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(1)

Since eq (1) passes through the point (0, -3) then

9 + 2f(-3) + c = 0

⇒ 9 – 6f + c = 0

⇒ -6f + c = -9 ……..(2)

Given equations of the circles are

x^{2} + y^{2} – 6x + 3y + 5 = 0 ………(3)

x^{2} + y^{2} – x – 7y = 0 ………(4)

Since the circles (1) & (3) are orthogonal then 2gg’ + 2ff’ = c + c’

⇒ 2g(-3) + 2f(\(\frac{3}{2}\)) = c + 5

⇒ -6g + 3f – c = 5 …….(5)

Since the circles (1) & (4) are orthogonal then 2gg’ + 2ff’ = c + c’

Question 5.

Find the equation of the circle passing through the origin having its centre on the line x + y = 4 and intersecting the circle x^{2} + y^{2} – 4x + 2y + 4 = 0 orthogonally.

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(1)

Since eq. (1) passes through the point (0, 0) then c = 0

Centre of (1), C = (-g, -f) lies on the line

x + y = 4 then -g – f = 4 ……..(2)

Given the equation of the circle is

x^{2} + y^{2} – 4x + 2y + 4 = 0 ……….(3)

Since the circles (1) & (3) are orthogonal then 2gg’ + 2ff’ = c + c’

2g(-2) + 2f(1) = c + 4

⇒ -4g + 2f = 4

⇒ -2g + f = 2 ……….(4)

Solve (2) & (4)

∴ The equation of the required circle is x^{2} + y^{2} + 2(-2)x + 2(-2)y + 0 = 0

⇒ x^{2} + y^{2} – 4x – 4y = 0

Question 6.

Find the equation of the circle passing through the points (2, 0), (0, 2) and orthogonal to the circle 2x^{2} + 2y^{2} + 5x – 6y + 4 = 0. [(TS) May ’19]

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(1)

Since, (1) passes through the point (2, 0), then

(2)^{2} + (0)^{2} + 2g(2) + 2f(0) + c = 0

⇒ 4 + 4g + c = 0

⇒ 4g + c = -4 ………(2)

Since (1) passes through the point (0, 2), then

(0)^{2} + (2)^{2} + 2g(0) + 2f(0) + c = 0

⇒ 4 + 4f + c = 0

⇒ 4f + c = -4 …….(3)

Given the equation of the circle is

2x^{2} + 2y^{2} + 5x – 6y + 4 = 0

⇒ x^{2} + y^{2} + \(\frac{5}{2}\)x – 3y + 2 = 0 ………(4)

Since (1) and (4) are orthogonal, then 2gg’ + 2ff’ = c + c’

2g(\(\frac{5}{4}\)) + 2f(\(\frac{-3}{2}\)) = c + 2

⇒ \(\frac{5g}{2}\) – 3f = c + 2

⇒ 5g – 6f = 2c + 4

⇒ 5g – 6f – 2c = 4 ………(5)

Question 7.

Find the equation of the circle which cuts orthogonally the circle x^{2} + y^{2} – 4x + 2y – 7 = 0 and having the centre at (2, 3). [Mar. ’19 (TS)]

Solution:

Given the equation of the circle is

x^{2} + y^{2} – 4x + 2y – 7 = 0 ……….(1)

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ………(2)

centre (-g, -f) = (2, 3)

∴ g = -2, f = -3

since (1) &(2) are orthogonal then 2gg’ + 2ff’ = c + c’

⇒ 2(-2)(-2) + 2(-3)(1) = -7 + c

⇒ 8 – 6 = -7 + c

⇒ 2 = -7 + c

⇒ c = 7 + 2

⇒ c = 9

∴ The equation of the required circle is x^{2} + y^{2} – 4x – 6y + 9 = 0

Question 8.

Find the equation of the circle which is orthogonal to each of the following three circles x^{2} + y^{2} + 2x + 17y + 4 = 0, x^{2} + y^{2} + 7x + 6y + 11 = 0, and x^{2} + y^{2} – x + 22y + 3 = 0. [May ’08; Mar. ’03]

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……..(1)

Given equations of the circles are

x^{2} + y^{2} + 2x + 17y + 4 = 0 ……..(2)

x^{2} + y^{2} + 7x + 6y + 11 = 0 ……..(3)

x^{2} + y^{2} – x + 22y + 3 = 0 ……….(4)

Since the circles (1) & (2) are orthogonal to each other then

2gg’ + 2ff’ = c + c’

⇒ 2g(1) + 2f(\(\frac{17}{2}\)) = c + 4

⇒ 2g + 17f = c + 4 ……….(5)

Since the circles (1) & (3) are orthogonal to each other then

2gg’ + 2ff’ = c + c’

⇒ 2g(\(\frac{7}{2}\)) + 2f(3) = c + 11

⇒ 7g + 6f – c = 11 ………(6)

Since the circles (1) & (4) are orthogonal to each other then

2gg’ + 2ff’ = c + c’

⇒ 2g(\(\frac{-1}{2}\)) + 2f(11) = c + 3

⇒ -g + 22f – c = 3 ………(7)

From (5) & (6)

Substitute g, f values in eq. (5)

2(-3) + 17(-2) – c = 4

⇒ -6 – 34 – c = 4

⇒ c = -44

∴ The equation of the required circle is x^{2} + y^{2} + 2(-3)x + 2(-2)y – 44 = 0

⇒ x^{2} + y^{2} – 6x – 4y – 44 = 0

Question 9.

Find the equation of the circle which cuts the circles x^{2} + y^{2} + 2x + 4y + 1 = 0, 2x^{2} + 2y^{2} + 6x + 8y – 3 = 0, x^{2} + y^{2} – 2x + 6y – 3 = 0 orthogonally.

Solution:

Given equations of the circles are

S = x^{2} + y^{2} + 2x + 4y + 1 = 0

S’ = 2x^{2} + 2y^{2} + 6x + 8y – 3 = 0

S’ = x^{2} + y^{2} + 3x + 4y – \(\frac{3}{2}\) = 0

S” = x^{2} + y^{2} – 2x + 6y – 3 = 0

The equation of the radical axis of the circles S = 0 and S’ = 0 are S – S’ = 0

⇒ x^{2} + y^{2} + 2x + 4y + 1 – (x^{2} + y^{2} + 3x + 4y – \(\frac{3}{2}\)) = 0

⇒ x^{2} + y^{2} + 2x + 4y + 1 – x^{2} – y^{2} – 3x – 4y + \(\frac{3}{2}\) = 0

⇒ -x + 1 + \(\frac{3}{2}\) = 0

⇒ -2x + 2 + 3 = 0

⇒ 2x – 5 = 0 ……..(1)

The equation of the radical axis of the circles S’ = 0 and S” = 0 is S’ – S” = 0

⇒ x^{2} + y^{2} + 3x + 4y – \(\frac{3}{2}\) – (x^{2} + y^{2} – 2x + 6y – 3) = 0

⇒ x^{2} + y^{2} + 3x + 4y – \(\frac{3}{2}\) – x^{2} – y^{2} + 2x – 6y + 3 = 0

⇒ 5x – 2y – \(\frac{3}{2}\) + 3 = 0

⇒ 10x – 4y – 3 + 6 = 0

⇒ 10x – 4y + 3 = 0 ……..(2)

Solving (1) and (2)

∴ Radical centre, C = (\(\frac{5}{2}\), 7) = centre of the required circle.

Radius of the required circle, r = the length of tangent from the radical centre, C = (\(\frac{5}{2}\), 7) of the circle is \(\sqrt{\mathrm{S}_{11}}\)

4x^{2} + 25 – 20x + 4y^{2} – 56y + 196 = 357

x^{2} + y^{2} – 5x – 14y – 34 = 0

Question 10.

Find the equation of the circle which intersects each of the following circles orthogonally.

x^{2} + y^{2} + 4x + 2y + 1 = 0, 2(x^{2} + y^{2}) + 8x + 6y – 3 = 0, x^{2} + y^{2} + 6x – 2y – 3 = 0

Solution:

Let, the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 ……(1)

Given equations of the circles are

x^{2} + y^{2} + 4x + 2y + 1 = 0 …….(2)

2(x^{2} + y^{2}) + 8x + 6y – 3 = 0

x^{2} + y^{2} + 4x + 3y – \(\frac{3}{2}\) = 0 ………(3)

x^{2} + y^{2} + 6x – 2y – 3 = 0 ……..(4)

Since (1) and (2) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(2) + 2f(1) = c + 1

⇒ 4g + 2f = c + 1

⇒ 4g + 2f – c = 1 ………(5)

Since (1) and (3) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(2) + 2f(\(\frac{3}{2}\)) = c – \(\frac{3}{2}\)

⇒ 4g + 3f = c – \(\frac{3}{2}\)

⇒ 4g + 3f – c = \(-\frac{3}{2}\) ………(6)

Since (1) and (4) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(3) + 2f(-1) = c – 3

⇒ 6g- 2f = c – 3

⇒ 6g – 2f – c = -3 ………(7)

From (5) and (6)

∴ The equation of the required circle is from (1)

x^{2} + y^{2} + 2(-7)x + 2(\(\frac{-5}{2}\))y – 34 = 0

⇒ x^{2} + y^{2} – 14x – 5y – 34 = 0

Question 11.

Find the equation of the circle which cuts the circles x^{2} + y^{2} – 4x – 6y + 11 = 0 and x^{2} + y^{2} – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [Mar. ’16 (AP); May ’07]

Solution:

Let the equation of the required circle is

x^{2} + y^{2} + 2gx + 2fy + c = 0 …….(1)

Given equations of the circles are

x^{2} + y^{2} – 4x – 6y + 11 = 0 ……..(2)

x^{2} + y^{2} – 10x – 4y + 21 = 0 ……..(3)

Since the circles (1) & (2) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(-2) + 2f(-3) = c + 11

⇒ -4g – 6f – c = 11 ……..(4)

Since the circles (1) & (3) are orthogonal then

2gg’ + 2ff’ = c + c’

⇒ 2g(-5) + 2f(-2) = c + 21

⇒ -10g – 4f – c = 21 ……..(5)

Centre of (1) is C = (-g, -f) lies on the line 2x + 3y = 7 then

2(-g) + 3(-f) = 7

⇒ -2g – 3f = 7 ……..(6)

From (4) & (5)

Substitute the values of g, f in eq. (4)

-4(-2) – 6(-1) – c = 11

⇒ 8 + 6 – c = 11

⇒ 14 – c = 11

⇒ c = 3

∴ The equation of the required circle is x^{2} + y^{2} + 2(-2)x + 2(-1)y + 3 = 0

⇒ x^{2} + y^{2} – 4x – 2y + 3 = 0

Question 12.

Find the equation of the circle passing through the points of intersection of the circles x^{2} + y^{2} – 8x – 6y + 21 = 0, x^{2} + y^{2} – 2x – 15 = 0, and (1, 2). [Mar. ’19 (AP); (TS) May ’17]

Solution:

Given equations of the circles are

S = x^{2} + y^{2} – 8x – 6y + 21 = 0

S’ = x^{2} + y^{2} – 2x – 15 = 0

Let, the given point A = (1, 2)

The equation of the circle passing through A, B is S + λS’ = 0

(x^{2} + y^{2} – 8x – 6y + 21) + λ(x^{2} + y^{2} – 2x – 15) = 0 ……….(1)

Since, (1) passes through the point (1, 2) then

((1)^{2} + (2)^{2} – 8(1) – 6(2) + 21) + λ[(1)^{2} + (2)^{2} – 2(1) – 15] = 0

⇒ 1 + 4 – 8 – 12 + 21 + λ(1 + 4 – 2 – 15) = 0

⇒ 6 + λ(-12) = 0

⇒ -12λ = -6

⇒ λ = \(\frac{1}{2}\)

∴ The equation of the required circle is

From (1)

⇒ x^{2} + y^{2} – 8x – 6y + 21 + \(\frac{1}{2}\)(x^{2} + y^{2} – 2x – 15) = 0

⇒ 2x^{2} + 2y^{2} – 16x – 12y + 42 + x^{2} + y^{2} – 2x – 15 = 0

⇒ 3x^{2} + 3y^{2} – 18x – 12y + 27 = 0

⇒ x^{2} + y^{2} – 6x – 4y + 9 = 0

Question 13.

Find the equation of the circle passing through the intersection of the circles x^{2} + y^{2} = 2ax and x^{2} + y^{2} = 2by and having its centre on the line \(\frac{x}{a}-\frac{\mathbf{y}}{b}\) = 2.

Solution:

Given equations of the circles are

S = x^{2} + y^{2} – 2ax = 0

S’ = x^{2} + y^{2} – 2by = 0

Given the equation of the straight line is

L = \(\frac{x}{a}-\frac{\mathbf{y}}{b}\) – 2 = 0

The equation of the circle passing through A, B is S + λS’ = 0

(x^{2} + y^{2} – 2ax) + λ(x^{2} + y^{2} – 2by) = 0 ……(1)

x^{2} + y^{2} – 2ax + λx^{2} + λy^{2} – 2λby = 0

(1 + λ) x^{2} + (1 + λ) y^{2} – 2ax – 2λby = 0

∴ The equation of the circle on \(\overline{\mathrm{AB}}\) as a diameter is from (1),

(x^{2} + y^{2} – 2ax) – \(\frac{1}{3}\) (x^{2} + y^{2} – 2by) = 0

⇒ 3x^{2} + 3y^{2} – 6ax – x^{2} – y^{2} + 2by = 0

⇒ 2x^{2} + 2y^{2} – 6ax + 2by = 0

⇒ x^{2} + y^{2} – 3ax + by = 0

Question 14.

If the straight line 2x + 3y = 1 intersects the circle x^{2} + y^{2} = 4 at points A and B then find the equation of the circle having \(\overline{\mathrm{AB}}\) as a diameter.

Solution:

Given equation of the circle is S = x^{2} + y^{2} – 4 = 0

Given the equation of the straight line is 2x + 3y – 1 = 0

The equation of the circle passing through A, B is S + λL = 0

x^{2} + y^{2} – 4 + λ(2x + 3y – 1) = 0 ………(1)

x^{2} + y^{2} – 4 + 2λx + 3λy – λ = 0

Here g = λ, f = \(\frac{3 \lambda}{2}\), c = -λ

Centre of the circle (1) is

C = (-g, -f) = (-λ, \(-\frac{3 \lambda}{2}\))

If \(\overline{\mathrm{AB}}\) is a diameter of a circle (1) then ‘C’ lies on 2x + 3y – 1 = 0

⇒ 2(-λ) + 3(\(-\frac{3 \lambda}{2}\)) – 1 = 0

⇒ -4λ – 9λ – 2 = 0

⇒ -13λ – 2 = 0

⇒ -13λ = 2

⇒ λ = \(\frac{-2}{13}\)

∴ The equation of the circle on \(\overline{\mathrm{AB}}\) as a diameter is, from (1)

⇒ x^{2} + y^{2} – \(\frac{2}{13}\) (2x + 3y – 1) = 0

⇒ 13x^{2} + 13y^{2} – 52 – 4x – 6y + 2 = 0

⇒ 13x^{2} + 13y^{2} – 4x – 6y – 50 = 0

Question 15.

If x + y = 3 is the equation of the chord AB of the circle x^{2} + y^{2} – 2x + 4y – 8 = 0. Find the equation of the circle having AB as the diameter. [(AP) May ’17, ’16, Mar. ’15]

Solution:

Given the equation of the circle is

S = x^{2} + y^{2} – 2x + 4y – 8 = 0

Given the equation of the straight line is

L = x + y – 3 = 0

The equation of the circle passing through A, B is S + λL = 0

(x^{2} + y^{2} – 2x + 4y – 8) + λ(x + y – 3) = 0 ……….(1)

x^{2} + y^{2} – 2x + 4y – 8 + λx + λy – 3λ = 0

x^{2} + y^{2} + (-2 + λ)x + (4 + λ)y – 8 – 3λ = 0

Here g = \(\frac{-2+\lambda}{2}\), f = \(\frac{4+\lambda}{2}\), c = -8 – 3λ

Centre of the circle (1) is

c = (-g, -f) = \(\left(\frac{2-\lambda}{2}, \frac{-4-\lambda}{2}\right)\)

If \(\overline{\mathrm{AB}}\) is diameter of circle (1), then ‘c’ lies on x + y – 3 = 0

⇒ \(\frac{2-\lambda}{2}+\frac{-4-\lambda}{2}-3=0\)

⇒ \(\frac{2-\lambda-4-\lambda-6}{2}=0\)

⇒ -8 – 2λ = 0

⇒ 2λ = -8

⇒ λ = -4

∴ The equation of the circle on \(\overline{\mathrm{AB}}\) as a diameter is from (1),

x^{2} + y^{2} – 2x + 4y – 8 – 4(x + y – 3) = 0

⇒ x^{2} + y^{2} – 2x + 4y – 8 – 4x – 4y + 12 = 0

⇒ x^{2} + y^{2} – 6x + 4 = 0

Question 16.

Find the radical centre of the three circles x^{2} + y^{2} – 4x – 6y + 5 = 0, x^{2} + y^{2} – 2x – 4y – 1 = 0, x^{2} + y^{2} – 6x – 2y = 0. [(AP) May ’19, Mar. ’18; (TS) May ’18]

Solution:

Given circles are

x^{2} + y^{2} – 4x – 6y + 5 = 0 …….(1)

x^{2} + y^{2} – 2x – 4y – 1 = 0 ……….(2)

x^{2} + y^{2} – 6x – 2y = 0 ……….(3)

The radical axis of (1) & (2) is

x^{2} + y^{2} – 4x – 6y + 5 – (x^{2} + y^{2} – 2x – 4y – 1) = 0

⇒ -2x – 2y + 6 = 0

⇒ x + y – 3 = 0 ……..(4)

The radical axis of (2) & (3) is

x^{2} + y^{2} – 2x – 4y – 1 – (x^{2} + y^{2} – 6x – 2y) = 0

⇒ 4x – 2y – 1 = 0 ……….(5)

Solving (4) & (5) we get

Question 17.

Find the radical centre of the circles x^{2} + y^{2} + 4x – 7 = 0, 2x^{2} + 2y^{2} + 3x + 5y – 9 = 0, x^{2} + y^{2} + y = 0. [May ’16 (TS) May ’14]

Solution:

Given equations of the circles are

S = x^{2} + y^{2} + 4x – 7 = 0

S’ = 2x^{2} + 2y^{2} + 3x + 5y – 9 = 0

S’ = \(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y-\frac{9}{2}=0\)

S” = x^{2} + y^{2} + y = 0

The equation of the radical axis of the circles S = 0 and S’ = 0 is S – S’ = 0

⇒ (x^{2} + y^{2} + 4x – 7) – (\(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y-\frac{9}{2}\)) = 0

⇒ x^{2} + y^{2} + 4x – 7 – x^{2} – y^{2} – \(\frac{3}{2} x-\frac{5}{2} y+\frac{9}{2}\) = 0

⇒ \(4 x-\frac{3}{2} x-7-\frac{5}{2} y+\frac{9}{2}=0\)

⇒ 8x – 3x – 14 – 5y + 9 = 0

⇒ 5x – 5y – 5 = 0

⇒ x – y – 1 = 0 ……..(1)

The equation of the radical axis of the circles S = 0 and S” = 0 is S – S” = 0

⇒ x^{2} + y^{2} + 4x – 7 – (x^{2} + y^{2} + y) = 0

⇒ x^{2} + y^{2} + 4x – 7 – x^{2} – y^{2} – y = 0

⇒ 4x – y – 7 = 0 ……….(2)

Solving (1) and (2)

∴ Radical centre = (2, 1)

Question 18.

Find the equation and length of the common chord of the two circles x^{2} + y^{2} + 3x + 5y + 4 = 0 and x^{2} + y^{2} + 5x + 3y + 4 = 0. [Mar. ’18 (TS)]

Solution:

Given circles are

x^{2} + y^{2} + 3x + 5y + 4 = 0 ………(1)

x^{2} + y^{2} + 5x + 3y + 4 = 0 ……….(2)

Equation of common chord of (1) & (2) as

x^{2} + y^{2} + 3x + 5y + 4 – (x^{2} + y^{2} + 5x + 3y + 4) = 0

⇒ -2x + 2y = 0

⇒ x – y = 0

The centre of 1st circle is c\(\left(\frac{-3}{2}, \frac{-5}{2}\right)\) and

Question 19.

Find the equation and length of the common chord of the circles x^{2} + y^{2} + 2x + 2y + 1 = 0, x^{2} + y^{2} + 4x + 3y + 2 = 0. [Mar. ’17 (AP & TS); May ’15 (AP)]

Solution:

Given circles are x^{2} + y^{2} + 2x + 2y + 1 = 0 …….(1)

and x^{2} + y^{2} + 4x + 3y + 2 = 0 ……..(2)

The common chord of (1) & (2) is

x^{2} + y^{2} + 2x + 2y + 1 – [x^{2} + y^{2} + 4x + 3y + 2] = 0

⇒ -2x – y – 1 = 0

⇒ 2x + y + 1 = 0

For the 1st circle x^{2} + y^{2} + 2x + 2y + 1 = 0

centre C = (-1, -1) and

radius r = \(\sqrt{1+1-1}\) = 1

d = perpendicular distance from (-1, -1) to the chord 2x + y + 1 = 0

Question 20.

Find the equation of the circle whose diameter is the common chord of the circles S = x^{2} + y^{2} + 2x + 3y + 1 = 0 and S’ = x^{2} + y^{2} + 4x + 3y + 2 = 0. (Apr. ’91)

Solution:

Given equations of the circles are,

S = x^{2} + y^{2} + 2x + 3y + 1 = 0

S’ = x^{2} + y^{2} + 4x + 3y + 2 = 0

The equation of the common chord of the circles S = 0 and S’ = 0 is S – S’ = 0

x^{2} + y^{2} + 2x + 3y + 1 – x^{2} – y^{2} – 4x – 3y – 2 = 0

2x + 1 = 0

L = 2x + 1 = 0

The equation of any circle passing through the point of intersection of S = 0 and L = 0 is S + λL = 0

(x^{2} + y^{2} + 2x + 3y + 1) + λ(2x + 1) = 0 ……..(1)

x^{2} + y^{2} + 2x + 3y + 1 + 2λx + λ = 0

x^{2} + y^{2} + 2(1 + λ)x + 3y + 1 + λ = 0

g = (1 + λ), f = \(\frac{3}{2}\)

Centre C = (-g, -f) = \(\left(-(1+\lambda), \frac{-3}{2}\right)\)

Since \(\overline{\mathrm{AB}}\) is a diameter of circle (1) then centre C(-1 – λ, \(\frac{-3}{2}\)) lies on the L = 0

2(-1 – λ) + 1 = 0

⇒ -2 – 2λ + 1 = 0

⇒ 2λ = -1

⇒ λ = \(\frac{-1}{2}\)

∴ The equation of the required circle is, from (1)

x^{2} + y^{2} + 2x + 3y + 1 – \(\frac{1}{2}\)(2x + 1) = 0

⇒ 2x^{2} + 2y^{2} + 4x + 6y + 2 – 2x – 1 = 0

⇒ 2x^{2} + 2y^{2} + 2x + 6y + 1 = 0

Question 21.

Show that the common chord of the circles x^{2} + y^{2} – 6x – 4y + 9 = 0, x^{2} + y^{2} – 8x – 6y + 23 = 0 is the diameter of the second circle, and also find its length.

Solution:

Given equations of the circles are

S = x^{2} + y^{2} – 6x – 4y + 9 = 0 ………(1)

S’ = x^{2} + y^{2} – 8x – 6y + 23 = 0 ………..(2)

The equation of the common chord of the circles S = 0 and S’ = 0 is S – S’ = 0

⇒ (x^{2} + y^{2} – 6x – 4y + 9) – (x^{2} + y^{2} – 8x – 6y + 23) = 0

⇒ x^{2} + y^{2} – 6x – 4y + 9 – x^{2} – y^{2} + 8x + 6y – 23 = 0

⇒ 2x + 2y – 14 = 0

⇒ x + y – 7 = 0 ……….(1)

Centre of the circle S’ = 0 is c = (-g, -f) = (4, 3)

The radius of the circle S’ = 0 is

r = \(\sqrt{(-4)^2+(-3)^2-23}\)

= \(\sqrt{16+9-23}\)

= √2

Now substituting the centre of the circle S’ = 0, c = (4, 3) in (1)

4 + 3 – 7 = 0

⇒ 7 – 7 = 0

⇒ 0 = 0

∴ (1) is the diameter of the second circle S’ = 0

Now, the length of the common chord = diameter of the second circle

= 2 × radius of the second circle

= 2r

= 2√2

Question 22.

Show that the circles x^{2} + y^{2} – 2x – 4y – 20 = 0 and x^{2} + y^{2} + 6x + 2y – 90 = 0 touch each other internally. Find the point of contact and the equation of common tangent. [(TS) Mar. ’15)]

Solution:

Given equations of the circles are

S = x^{2} + y^{2} – 2x – 4y – 20 = 0 …….(1)

S’ = x^{2} + y^{2} + 6x + 2y – 90 = 0 …….(2)

for the circle (1)

Centre C_{1} = (1, 2)

|r_{1} – r_{2}| = |5 – 10| = |-5| = 5

∴ c_{1}c_{2} = |r_{1} – r_{2}|

∴ Given circles touch each other internally.

Let ‘P’ be the point of contact.

Now ‘P’ divides C_{1}C_{2} in the ratio r_{1} : r_{2} (5 : 10 = 1 : 2) externally.

∴ Point of contact

In this case, the common tangent is nothing but the radical axis.

∴ The equation of the common tangent (radical axis) of the circles S = 0 and S’ = 0 is S – S’ = 0

(x^{2} + y^{2} – 2x – 4y – 20) – (x^{2} + y^{2} + 6x + 2y – 90) = 0

⇒ x^{2} + y^{2} – 2x – 4y – 20 – x^{2} – y^{2} – 6x – 2y + 90 = 0

⇒ -8x – 6y + 70 = 0

⇒ 4x + 3y – 35 = 0

Question 23.

Show that the circles x^{2} + y^{2} – 8x – 2y + 8 = 0 and x^{2} + y^{2} – 2x + 6y + 6 = 0 touch each other and find the point of contact. (Mar. ’14)

Solution:

Given equations of the circles are

S = x^{2} + y^{2} – 8x – 2y + 8 = 0 ………(1)

S’ = x^{2} + y^{2} – 2x + 6y + 6 = 0 ……..(2)

For circle (1),

centre C_{1} = (-g, -f) = (4, 1)

r_{1} + r_{2} = 3 + 2 = 5

∴ C_{1}C_{2} = r_{1} + r_{2}

∴ The given circles touch other externally.

The point of contact, P divides C_{1}C_{2} internally in the ratio (r_{1} : r_{2} = 3 : 2)

Question 24.

Show that the circles x^{2} + y^{2} – 2x = 0 and x^{2} + y^{2} + 6x – 6y + 2 = 0 touch each other. Find the coordinates of the point of contact. Is the point of contact external or internal?

Solution:

Given equations of the circles are

x^{2} + y^{2} – 2x = 0 ………(1)

x^{2} + y^{2} + 6x – 6y + 2 = 0 ……….(2)

For circle (1), centre, C_{1} = (1, 0)

r_{1} + r_{2} = 1 + 4 = 5

∴ C_{1}C_{2} = r_{1} + r_{2}

∴ Given circles touch each other externally.

Let ‘P’ be the point of contact.

Now, the point of contact, ‘P’ divides C_{1}C_{2} in the ratio r_{1} : r_{2} (1 : 4) internally.

∴ The point of contact is internal.

Question 25.

Show that the circles x^{2} + y^{2} + 2ax + c = 0 and x^{2} + y^{2} + 2by + c = 0 touch each other if \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c}\).

Solution:

Given equations of the circles are

x^{2} + y^{2} + 2ax + c = 0 ………(1)

x^{2} + y^{2} + 2by + c = 0 ………(2)

Centre of (1) is C_{1} = (-a, 0)

Radius of (1) is r_{1} = \(\sqrt{a^2-c}\)

Centre of (2) is C_{2} = (0, -b)

Radius of (2) is r_{2} = \(\sqrt{b^2-c}\)

C_{1}C_{2} = \(\sqrt{(-a+0)^2+(0+b)^2}=\sqrt{a^2+b^2}\)

Since the given two circles touch each other then

C_{1}C_{2} = |r1 ± r2|

\(\sqrt{a^2+b^2}=\left|\sqrt{a^2-c} \pm \sqrt{b^2-c}\right|\)

Squaring on both sides

Question 26.

If the straight line represented by x cos α + y sin α = p intersects the circle x^{2} + y^{2} = a^{2} at the points A and B then show that the equation of the circle with \(\overline{\mathbf{A B}}\) as the diameter is x^{2} + y^{2} – a^{2} – 2p(x cos α + y sin α – p) = 0.

Solution:

Given the equation of the circle is

S = x^{2} + y^{2} – a^{2} = 0

Given the equation of the straight line is

L = x cos α + y sin α – p = 0

The equation of the circle passing through A, B is S + λL = 0

(x^{2} + y^{2} – a^{2}) + λ(x cos α + y sin α – p) = 0 ……(1)

x^{2} + y^{2} – a^{2} + λx cos α + λy sin α – λp = 0

x^{2} + y^{2} + λ cos αx + λ sin αy – a^{2} – λp = 0

Here,

The equation of the circle on \(\overline{\mathbf{A B}}\) as diameter is from (1)

(x^{2} + y^{2} – a^{2}) – 2p(x cos α + y sin α – p) = 0

Question 27.

If the two circles x^{2} + y^{2} + 2gx + 2fy = 0 and x^{2} + y^{2} + 2g’x + 2f’y = 0 touch each other, then show that f’g = fg’. [(AP) Mar. ’20, (TS) ’16]

Solution:

Given equations of the circles are

x^{2} + y^{2} + 2gx + 2fy = 0 ……….(1)

x^{2} + y^{2} + 2g’x + 2f’y = 0 ……..(2)

For circle (1),

Center, C_{1} = (-g, -f)

Radius, r_{1} = \(\sqrt{\mathrm{g}^2+\mathrm{f}^2}\)

For circle (2),

Center, C_{2} = (-g’, -f’)

radius, r_{2} = \(\sqrt{g^{\prime 2}+f^{\prime 2}}\)

Now, C_{1}C_{2} = \(\sqrt{\left(-\mathrm{g}+\mathrm{g}^{\prime}\right)^2+\left(-\mathrm{f}+\mathrm{f}^{\prime}\right)^2}\)

Since the given circles touch each other then

Question 28.

Find the equation of a circle that cuts each of the following circles orthogonally. (Mar. ’07)

S’ = x^{2} + y^{2} + 3x + 2y + 1 = 0,

S” = x^{2} + y^{2} – x + 6y + 5 = 0,

S'” = x^{2} + y^{2} + 5x – 8y + 15 = 0

Solution:

Given equations of the circles are

S’ = x^{2} + y^{2} + 3x + 2y + 1 = 0

S” = x^{2} + y^{2} – x + 6y + 5 = 0

S'” = x^{2} + y^{2} + 5x – 8y + 15 = 0

The equation of the radical axis of S’ = 0 and S” = 0 is S’ – S” = 0

⇒ x^{2} + y^{2} + 3x + 2y + 1 – x^{2} – y^{2} + x – 6y – 5 = 0

⇒ 4x – 4y – 4 = 0

⇒ x – y – 1 = 0 ……….(1)

The equation of the radical axis of S” = 0 and S'” = 0 is S” – S'” = 0

⇒ x^{2} + y^{2} – x + 6y + 5 – x^{2} – y^{2} – 5x + 8y – 15 = 0

⇒ -6x + 14y – 10 = 0

⇒ -3x + 7y – 5 = 0

⇒ 3x – 7y + 5 = 0 ……….(2)

Solving (1) & (2)

∴ Radical centre = (3, 2) centre of the required circle

Radius = The length of the tangent from (3, 2) to the circle (1) is \(\sqrt{\mathrm{S}_{11}}\)

The equation of the required circle is (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – 3)^{2} + (y – 2)^{2} = 27

⇒ x^{2} + 9 – 6x + y^{2} + 4 – 4y – 27 = 0

⇒ x^{2} + y^{2} – 6x – 4y – 14 = 0

Question 29.

Prove that the radical axis of any two circles is perpendicular to the line joining their centres.

Solution:

Let S = x^{2} + y^{2} + 2gx + 2fy + c = 0

S’ = x^{2} + y^{2} + 2g’x + 2f’y + c’ = 0 be the given circles.

The equation of the radical axis is S – S’ = 0

(x^{2} + y^{2} + 2gx + 2fy + c) – (x^{2} + y^{2} + 2g’x + 2f’y + c’) = 0

⇒ x^{2} + y^{2} + 2gx + 2fy + c – x^{2} – y^{2} – 2g’x – 2f’y – c’ = 0

⇒ 2(g – g’)x + 2(f – f’)y + c – c’ = 0

The slope of the radical axis is

∴ The radical axis is perpendicular to the line of centres.

Question 30.

Prove that the radical axis of the circle x^{2} + y^{2} + 2gx + 2fy + c = 0 and x^{2} + y^{2} + 2g’x + 2f’y + c’ = 0 is the diameter of the latter circle (or the former bisects the circumference of the latter) if 2g'(g – g’) + 2f'(f – f’) = c – c’.

Solution:

Given equations of the circles are

S = x^{2} + y^{2} + 2gx + 2fy + c = 0

S’ = x^{2} + y^{2} + 2g’x + 2f’y + c = 0

The equation of the radical axis of the circles S = 0 and S’ = 0 is S – S’ = 0

(x^{2} + y^{2} + 2gx – 2fy + c) – (x^{2} + y^{2} + 2g’x + 2f’y + c’) = 0

⇒ x^{2} + y^{2} + 2gx – 2fy + c – x^{2} – y^{2} – 2g’x – 2f’y – c’ = 0

⇒ 2(g – g’)x+2(f – f’)y + c – c’ = 0 ……(1)

Centre of the second circle S’ = 0 is c = (-g’, -f’)

Since (1) is the diameter of the second circle S’ = 0, then

substitute the centre of the second circle S’ = 0 is c = (-g’, -f’) in (1)

⇒ 2(g – g’)(-g’) + 2(f – f’)(-f’) + c – c’ = 0

⇒ 2g'(g – g’) + 2f(f – f’) = c – c’

Question 31.

If P and Q are conjugate with respect to a circle S = x^{2} + y^{2} + 2gx + 2fy + c = 0, then prove that the circle on PQ as diameter cuts the circle S = 0 orthogonally.

Solution:

Let P = (x_{1}, y_{1}) and Q = (x_{2}, y_{2}),

S = x^{2} + y^{2} + 2gx + 2fy + c = 0 ……….(1)

Given that, the two points P, Q are conjugate w.r.t a circle S = 0, then S_{12} = 0

⇒ x_{1}x_{2} + y_{1}y_{2} + g(x_{1} + x_{2}) + f(y_{1} + y_{2}) + c = 0 ………(2)

The equation of the circle having \(\overline{\mathrm{PQ}}\) as diameter is

S’ = (x – x_{1})(x – x_{2}) + (y – y_{1})(y – y_{2}) = 0

⇒ x^{2} – x_{1}x_{2} – xx_{1} + x_{1}x_{2} + y^{2} – yy_{2} – yy_{1} + y_{1}y_{2} = 0

⇒ x^{2} + y^{2} – x(x_{1} + x_{2}) – y(y_{1} + y_{2}) + x_{1}x_{2} + y_{1}y_{2} = 0 ………(3)

Now, applying orthogonally condition S = 0 and S’ = 0

2gg’ + 2ff’ = \(2 \mathrm{~g}\left(\frac{-\left(\mathrm{x}_1+\mathrm{x}_2\right)}{2}\right)+2 \mathrm{f}\left(\frac{-\left(\mathrm{y}_1+\mathrm{y}_2\right)}{2}\right)\)

= -g(x_{1} + x_{2}) – f(y_{1} + y_{2})

= x_{1}x_{2} + y_{1}y_{2} + c (∵ from(2))

c + c’ = c + x_{1}x_{2} + y_{1}y_{2}

= x_{1}x_{2} + y_{1}y_{2} + c

∴ 2gg’ + 2ff’ = c + c’

∴ The circles S = 0 and S’ = 0 are cut orthogonally.