Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B System of Circles Solutions Exercise 2(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B System of Circles Solutions Exercise 2(a)

I.

Question 1.

Find k if the following pairs of circles are orthogonal.

i) x² + y² + 2by – k = 0, x² + y² + 2ax + 8=0.

Solution:

g_{1} = 0; f_{1} = b; c_{1} = -k

g_{2} = a; f_{2} = 0; c_{2} = 8

Two circles are said to be orthogonal

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2(0)(a) +2(b)(0)= -k + 8

0 = – k + 8

k = 8

ii) x² + y² – 6x – 8y + 12 = 0;

x² + y² – 4x + 6y + k = 0

Solution:

g_{1} = -3; f_{1} = -4; c_{1} = 12

g_{2} = -2; f_{2} = 3; c_{2} = k

Two circles are said to be orthogonal

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2(-3)(-2) + 2(3)(-4) = 12 + k

+ 12 – 24 = 12+ k ⇒ k = -24

iii) x² + y² – 5x – 14y – 34 = 0,

x² + y² + 2x + 4y + k = 0

Solution:

g_{1} = \(\frac{-5}{2}\) ; f_{1} = -7 ; c_{1} = -34

g_{2} = 1; f_{2} = 2; c_{2} = k

Two circles are said to be orthogonal

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2(\(\frac{-5}{2}\)(1) + 2(-7)(2) = -34 + k

-5 – 28 = -34 + k

– 33 = – 34 + k

k = 34 – 33

⇒ k = 1

iv) x² + y² + 4x + 8 = 0, x² + y² – 16y +k = 0

Solution:

g_{1} =2 ; f_{1} = 0 ; c_{1} = 8

g_{2} = 0 ; f_{2} = – 4; c_{2} = k

Two circles are said to be orthogonal

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2(2)(0) + 2(0)(-8) = 8 + k

0 + 0 = 8 + k

⇒ k = -8

Question 2.

Find the angle between the circles given by the equations.

i) x² + y² – 12x – 6y + 41 = 0,

x² + y² + 4x + 6y – 59= 0.

Solution:

ii) x² + y²+ 6x – 10y – 135 = 0, x² + y² – 4x + 14y – 116 = 0.

Solution:

Question 3.

Show that the angle between the circles x + y = a , x + y = ax + ay is \(\frac{3 \pi}{4}\).

Solution:

Equations of the circles are

S ≡ x² + y² – a² = 0

S’ ≡ x² + y² – ax – ay = 0

Question 4.

Show that the circles given by the following equations intersect each other orthogonally.

i) x² + y² – 2x – 2y – 7 = 0,

3x² + 3y² – 8x + 29y = 0.

Solution:

C_{1} = (1, 1)

g = -1, f = -1, c =-7

g’ = \(\frac{-4}{3}\), f’ = \(\frac{29}{6}\) ; c’ = o

Condition that two circles are orthogonal is

∴ -7 = -7

Hence both circles cut orthogonally.

ii) x² + y² + 4x – 2y – 11 = 0,

x² + y² – 4x – 8y + 11 =0.

Solution:

g_{1} = 2, f_{1} = -1, c_{1} = -11

g_{2} = -2, f_{2} = -4, c_{2} = 11

Two circles are said to be orthogonal

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2(2)(-2) + 2(-1)(-4) = -11 + 11

-8 + 8 = 0

∴ Two circles are orthogonal.

iii) x² + y² – 2x + 4y + 4 = 0, .

x² + y² + 3x + 4y + 1 = 0.

Solution:

g = -1, f = 2, c = 4

g’ = \(\frac{3}{2}\), f’= 2, c’= 1

2gg’ + 2ff’ = c + c’

2(-1). \(\frac{3}{2}\) + 2 × 2 × 2 = 4 + 1

-3 + 8 = 5

5 = 5

Hence circles cut orthogonally.

iv) x² + y² – 2lx + g = 0, x² + y² + 2my – g = 0.

Solution:

g_{1} = -l ; f_{1} = 0, c_{1} = g, g_{2} = 0, f_{2} = m, c_{2}= -g

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2} is condition for two circles be orthogonal

2(-1)(0) + 2(0)(m) = g – g

0 = 0

∴ Two circles are orthogonal.

II.

Question 1.

Find the equation of the circle which passes through the origin and intersects the circles below, orthogonally.

i) x² + y² – 4x + 6y + 10 = 0, x² + y² + 12y + 6 = 0.

Solution:

Let equation of circle be

x² + y² + 2gx + 2fy + c = 0 ……… (i)

Above circle passes through origin

∴ c = 0

Circle (i) is orthogonal to

x² + y² – 4x + 6y + 10 = 0 then

2g(-2) + 2f(3) = 0 + 10

-4g + 6f = 10 ………… (ii)

Circle (i) is orthogonal to

x² + y² + 12y + 6 = 0

2g(0) + 2f(6) = 6 + 0

12f = 6

f = \(\frac{1}{2}\) …………… (iii)

Solving (ii) and (iii) we get

– 4g + 6 × \(\frac{1}{2}\) = 10

-4g = 10 – 3

g = –\(\frac{7}{4}\)

∴ Equation of circle be

x² + y² – \(\frac{7}{2}\)x + y = 0

2x² + 2y² – 7x + 2y = 0.

ii) x² + y² – 4x – 6y – 3 = 0, x² + y² – 8y + 12 = 0.

Solution:

Let the equation of the circle be

x² + y² + 2gx + 2fy + c = 0

It cuts

x² + y² – 4x – 6y – 3 = 0; x² + y² – 8y + 12 = 0

g_{1} = -2, f_{1} = -3, c_{1} = -3

g_{2} = 0 ; f_{2} = -4, c_{2} = 12

Let g,f, c be constants of required circle.

Required circle passes through origin

∴ c = 0

Requited circle is orthogonal to both circles.

∴ 2g(-2) +2f(-3) = -3 + 0 …………… (i)

2g(0) + 2f(-4) = 12 + 0 …………….. (ii)

Solving (i) and (ii) we get

2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

Condition of orthogonality

f = –\(\frac{3}{2}\), g = \(\frac{6}{2}\)

Required circle be x² + y² + 6x – 3y = 0

Question 2.

Find the equation of the circle which passes through the point (0, -3) and intersects the circles given by the equations x² + y² – 6x + 3y + 5 = 0 and x² + y² – x – 7y = 0 orthogonally.

Solution:

Let circle be x² + y² + 2gx + 2fy + c = 0 ……… (i)

(i) is orthogonal with x² + y² – 6x + 3y + 5 = 0

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

-6g + 3f = c + 5 ……… (ii)

(i) is orthogonal with x² + y² – x – 7y = 0

-g – 7f = c ……… (iii)

Circle passes through (0, -3)

0 + 9 – 6f + c = 0 ……… (iv)

(iii) – (ii)

5g – 10f = -5

g – 2f = -1

(iii) + (iv)

x² + y² + \(\frac{4}{3}\)y + \(\frac{2}{3}\)x – 5 = 0

3x² + 3y² + 4y + 2x – 15 = 0

(or) 3x² + 3y² + 2x + 4y – 15 = 0

Question 3.

Find the equation of the circle passing through the origin, having its centre on the line x + y = 4 and intersecting the circle x² + y² – 4x + 2y + 4 = 0 orthogonally.

Solution:

Let circle be x² + y²+ 2gx + 2fy + c = 0

Equation is passing through (0, 0)

0 + 0 + 2g.0 + 2f.0 + c = 0 ⇒ c = 0

x² + y² + 2gx + 2fy = 0

Centre passes through x + y = 4

∴ -g – f = 4 ………. (i)

Circle is orthogonal to

x² + y² – 4x + 2y + 4 = 0

-4g + 2f = 4 + 0

f – 2g = 2 ……… (ii)

Solving (i) and (ii) we get

– 3g = 6

9 = -2

f = -2

Equation of circle be x² + y² – 4x – 4y = 0

Question 4.

Find the equation of the circle which passes through the points (2, 0), (0, 2) and orthogonal to the circle 2x² + 2y² + 5x – 6y + 4 = 0.

Solution:

Let equation of circle be

x² + y² + 2gx + 2fy + c = 0

Passes through (2, 0), (0, 2) then

Orthogonal to x + y + \(\frac{5}{2}\)x – \(\frac{6}{2}\)y + 2 = 0

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

\(\frac{5}{2}\)g – 3f = 2 + c

But g = f

\(\frac{5}{2}\)g – 3g = 2 + c

⇒ -g = 4 + 2C

Putting value of g in equation (i)

-16 – 8c + c = -4

c = –\(\frac{12}{7}\)

– g = 4 – \(\frac{24}{7}\) = +\(\frac{4}{7}\)

g = \(\frac{-4}{7}\) = f

Equation of the circle is

x + y – \(\frac{8x}{7}\) – \(\frac{8y}{7}\) – \(\frac{12}{7}\) = 0

⇒ 7(x² + y²) – 8x – 8y – 12 = 0

Question 5.

Find the equation of the circle which cuts orthogonally the circle x² + y² – 4x + 2y- 7 = 0 and having a centre at (2, 3).

Solution:

The given circle is

x² + y² – 4x + 2y – 7 = 0 ……. (1)

Let the required circle which, cuts orthogonally the circle (1) is

x² + y² + 2gx + 2fy + c = 0 ………… (2)

Its centre is (-g, -f) = (2, 3) given

g = -2, f = -3

The two circles (1) and (2) are cutting each other orthogonally.

So 2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

2(-2)(-2) + 2(-3)(1) = – 7 + c

8 – 6 = -7 + c

+ 2 = -7 + c

c = 7 + 2 = 9 ⇒ c = 9

Hence the required circle is,

x² + y² – 4x – 6y + 9 = 0

III.

Question 1.

Find the equation of the circle which intersects the circle

x² + y² – 6x + 4y – 3 = 0 orthogonally and passes through the point (3, 0) and touches Y-axis.

Solution:

Let circle be (x – h)² + (y – k)² = r²

If circle touches Y-axis then co-ordinate of centre (h, k); radius = |h|

(x – h)² + (y – k)² = h²

x² – 2hx + h² + y²

-2ky + k² = h²

Orthogonal to

x² + y² – 6x + 4y – 3 = 0

2(-h) (-3) + 2(-k) (2) = -3 + k²

6h – 4k = -3 + k²

x² – 2hx + y²

-2ky + k² = h

Passes through (3, 0) 6h – 4k + 3 – k² = 0

9 – 6h + k² = 0 ……… (ii)

Adding .(i) and (ii) we get

c = 9

12 – 4k = 0 or k = 3, h = 3,

Equation of circle be y² + x² – 6x – 6y + 9 =0.

Question 2.

Find the equation of the circle which cuts the circles x² + y² – 4x – 6y + 11 =0 and x² + y² – 10x – 4y + 2t = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7.

Solution:

Let circle be x² + y² + 2gx + 2fy + c = 0 ………. (i) Orthogonal to circle

2g(-2) + 2f(-3) = 11 + c ……… (ii)

2g (-5) + 2f(-2) = 21 + c ……….. (iii)

Subtracting it we get

-6g + 2f = 10 ………. (iv)

Circles centre is on 2x + 3y = 7

∴ -2g – 3f = 7 …….. (v)

Solving (iv) and (v)

f = -1, g = -2, c = 3

Equation of circle be x² + y² – 4x – 2y + 3 = 0

Question 3.

If P, Q are conjugate points with respect to a circle S ≡ x² + y² + 2gx + 2fy + c = 0 then prove that the circle PQ as diameter cuts the circles S = 0 orthogonally.

Solution:

Let P = (x_{1}, y_{1}) and Q(x_{2}, y_{2}) be the conjugate points w.r.t. the circle

S ≡ x² + y² – a² = 0 …….. (i)

Polar of P w.r.t.(1) is xx_{1} + yy_{1} – a² = 0 ………… (ii)

Given P and Q are conjugate points ⇒ Q lies on (ii)

x_{1}x_{2} + y_{1}y_{2} – a² = 0 …….. (iii)

Equation of the circle on PQ as diameter is

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

⇒ x² + y² – (x_{1} + x_{2}) x – (y_{1} + y_{2})y + (x_{1}x_{2} + y_{1}y_{2}) = 0 …………… (iv)

(i) and (iv) are orthogonal.

2g_{1}g_{2} + 2f_{1}f_{2} = 2[0(\(\frac{-x_{1}+x_{2}}{2}\)) + 0(\(\frac{-y_{1}+y_{2}}{2}\))]

c_{1} + c_{2} = -a² + a²

⇒ 2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2}

∴ The circle on PQ as diameter cuts S orthogonally.

Question 4.

If the equations of two circles whose radii are a, a’ are S = 0 and S’ = 0, then show that the circles \(\frac{S}{a}+\frac{S’}{a’}\) = 0 and \(\frac{S}{a}-\frac{S’}{a’}\) = 0 intersect orthogonally.

Solution:

Let 2d be the distance between the centres of the circles S = 0 and S’ =. 0. Take the line joining the centres as X-axis and the point midway between the centres as origin.

Then the equations of the circles are

S ≡ (x – d)² + y² – a² = 0

S’ ≡ (x + d)² + y² – a² = 0

∴ \(\frac{S}{a}+\frac{S’}{a’}\) =0 becomes Sa’ + S’a = 0

(or) [(x – d)² + y² -a²]a’ + [(x + d)² + y² – a’²]a = 0

(or) x² + y² + 2{\(\frac{(a-a’)d}{(a+a’)}\)}x + (d² – aa’) = 0 ………… (i)

Putting -a’ for a’ in we get \(\frac{S}{a}-\frac{S’}{a’}\) = 0

reduces to x² + y² + 2{\(\frac{(a+a’)d}{(a-a’)}\)}x + (d² + aa’) = 0 ………… (ii)

If (i) and (ii) cut orthogonally then

L.H.S. = 2gg’ + 2ff = c + c’

= 2{\(\frac{(a+a’)d}{(a-a’)}\)}{\(\frac{(a+a’)d}{(a-a’)}\)} + 2(0). (0) = 2d²

= (d² – aa’) + (d² + aa’) = 2d²

Which being true the circles (i) and (ii) the circles \(\frac{S}{a}±\frac{S’}{a’}\) = 0 cut each other at right angles.

Question 5.

Find the equation of the circle which intersects each of the following circles orthognally

i) x² + y² + 2x + 4y + 1 = 0, x² + y² – 2x + 6y – 3 = 0, 2(x² + y²) +6x + 8y – 3 = 0.

Solution:

Let equation of circle be

x² + y² + 2gx + 2fy + c = 0

Given circle is orthogonal to all 3 circles then

2g(1) + 2f(2) = c + 1 ……….. (i)

2g(\(\frac{3}{2}\)) = 2f(2) = c – \(\frac{3}{2}\) ………… (ii)

2g (-1) +2f(3) = c – 3 ………… (iii)

(iii) – (ii)

-5g + 2f = \(\frac{-3}{2}\) (or) – 10g + 4f = -3 ……….. (iv)

(iii) – (i)

-4g + 2f = – 4

f – 2g = -2 ……….. (v)

Solving (iv) and (v) we get

f = -7, g = -5/2, c = -34

∴ Equation of circle be

x² + y² – 5x – 14y – 34 = 0

ii) x² + y² + 4x + 2y + 1 = 0, 2(x² + y²) + 8x + 6y – 3 = 0, x² + y² + 6x – 2y – 3 = 0.

Solution:

Let required circle equation be

x² + y² + 2gx + 2fy + c = 0 this circle is orthogonal to above three circles.

∴ 2g(2) + 2f(1) = c + 1 ……….. (i)

2g(2) + 2f(\(\frac{3}{2}\)) = c – \(\frac{3}{2}\) ……….. (ii)

2g(3) + 2f(-1) = c – 3 ……….. (iii)

(i) – (ii) we get (ii) – (iii) we get

– f = \(\frac{5}{2}\) then – 2g + 5f = \(\frac{3}{2}\)

We get

g = -7 and f = \(\frac{-5}{2}\)

Substituting ‘g’ and ‘f’ in (i) we get

4(-7) + 2(\(\frac{-5}{2}\)) = c + 1

c = -34

Required equation of circle be

x² +y² – 5y – 14x – 34 = 0

Question 6.

If the straight line 2x + 3y = 1 intersects the circle x² + y² = 4 at the points A and B, then find the equation of the circle having AB as diameter.

Solution:

Equation of circle passing through x² + y² – 4 and 2x + 3y – 1 =0 can be written as

(x² + y² – 4) + λ(2x + 3y – 1) = 0

x² + y² + 2λx + 3λy – 4 – λ = 0

Centre : (-λ, \(\frac{-3 \lambda}{2}\))

Centre lies on 2x + 3y – 1 =0

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

λ = \(\frac{-2}{13}\)

∴ Equation of circle be

13 (x² + y²) – 4 x 13 – 2(2x + 3y- 1) = 0

13 (x² + y²) – 4x – 6y – 50 = 0.

Question 7.

If x + y = 3 is the equation of the chord AB of the circle x² + y² – 2x + 4y – 8 = 0, find the equation of the circle having \(\overline{\mathrm{AB}}\) as diameter.

Solution:

Required equation of circle passing through intersection S = 0 and L = 0 is S + λL = 0

(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0

(x² + y² + x(-2 + λ) + y (4 + λ) – 8 – 3λ = 0 ………… (i)

x² + y² + 2gx + 2fy + c = 0 ………… (ii)

Comparing (i) and (ii) we get

2 – λ – 4 – λ = 6

-2λ = 8 ⇒ λ = -4

Required equation of circle be

(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0

x² + y² – 6x + 4 = 0

Question 8.

Find the equation of the circle passing through the intersection of the circles x² + y² = 2ax and x² + y² = 2by and having its centre on the line \(\frac{x}{a}-\frac{y}{b}\) = 2.

Solution:

Equation of circle passes through

x² + y² – 2ax = 0 and x² + y² – 2by = 0 can be written as

x² + y² – 2ax + X (x² + y² – 2by) = 0

x²(1 + λ) + y²(1 + λ) + x(-2a) – (2bλ)y = 0

1- λ = (1 + λ)

λ = -1/3

Equation of circle be

3x² + 3y² – 6ax – x² – y² + 2by = 0

⇒ 2x² + 2y² – 6ax + 2by = 0

⇒ x² + y² – 3ax + by = 0