Students must practice this TS Intermediate Maths 2B Solutions Chapter 2 System of Circles Ex 2(b) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2B Solutions Chapter 2 System of Circles Exercise 2(b)

I.

Question 1.

Find the equation of the radical axis of the following circles.

(i) x^{2} + y^{2} – 3x – 4y + 5 = 0, 3(x^{2} + y^{2}) – 7x + 8y – 11 = 0

Solution:

Let S ≡ x^{2} + y^{2} – 3x – 4y + 5 = 0 and S’ ≡ \(x^2+y^2-\frac{7}{3} x+\frac{8}{3} y-\frac{11}{3}=0\) be the given circles.

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

⇒ (x^{2} + y^{2} – 3x – 4y + 5) – (\(x^2+y^2-\frac{7}{3} x+\frac{8}{3} y-\frac{11}{3}\)) = 0

⇒ -3x – 4y + 5 + \(\frac{7}{3} x-\frac{8}{3} y+\frac{11}{3}\) = 0

⇒ -9x – 12y + 15 + 7x – 8y + 11 = 0

⇒ -2x – 20y + 26 = 0

⇒ x + 10y – 13 = 0

(ii) x^{2} + y^{2} + 2x + 4y + 1 = 0, x^{2} + y^{2} + 4x + y = 0

Solution:

Let S ≡ x^{2} + y^{2} + 2x + 4y + 1 = 0 and S’ ≡ x^{2} + y^{2} + 4x + y = 0 be the given circles.

Then the radical axis of circles S = 0, S’ = 0 is S – S’ = 0.

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

⇒ -2x + 3y + 1 = 0

⇒ 2x – 3y – 1 = 0

∴ The equation of the radical axis of circles is 2x – 3y – 1 = 0.

(iii) x^{2} + y^{2} + 4x + 6y – 7 = 0, 4(x^{2} + y^{2}) + 8x + 12y – 9 = 0.

Solution:

Let S ≡ x^{2} + y^{2} + 4x + 6y – 7 = 0 and S’ ≡ x^{2} + y^{2} + 2x + 3y – \(\frac{9}{4}\) = 0

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

⇒ 2x + 3y – \(\frac{19}{4}\) = 0

⇒ 8x + 12y – 19 = 0

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

Solution:

Let S ≡ x^{2} + y^{2} – 2x – 4y – 1 = 0 and S’ ≡ x^{2} + y^{2} – 4x – 6y + 5 = 0

∴ The equation of radical axis of S = 0, S’ = 0 is S – S’ = 0

⇒ 2x + 2y – 6 = 0

⇒ x + y – 3 = 0

Question 2.

Find the equation of the common chord of the following pair of circles.

(i) x^{2} + y^{2} – 4x – 4y + 3 = 0, x^{2} + y^{2} – 5x – 6y + 4 = 0

Solution:

Let S ≡ x^{2} + y^{2} – 4x – 4y + 3 = 0 and S’ = x^{2} + y^{2} – 5x – 6y + 4 = 0

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

⇒ x + 2y – 1 = 0

(ii) x^{2} + y^{2} + 2x + 3y + 1 = 0, x^{2} + y^{2} + 4x + 3y + 2 = 0.

Solution:

Denote S ≡ x^{2} + y^{2} + 2x + 3y + 1 = 0 and S’ ≡ x^{2} + y^{2} + 4x + 3y + 2 = 0

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

⇒ -2x – 1 = 0

⇒ 2x + 1 = 0

(iii) (x – a)^{2} + (y – b)^{2} = c^{2}, (x – b)^{2} + (y – a)^{2} = c^{2} (a ≠ b)

Solution:

Denote S ≡ (x – a)^{2} + (y – b)^{2} = c^{2}

⇒ x^{2} + y^{2} – 2ax – 2by + a^{2} + b^{2} – c^{2} = 0 …….(1)

and S’ ≡ (x – b)^{2} + (y – a)^{2} = c^{2}

⇒ x^{2} + y^{2} – 2bx – 2ay + b^{2} + a^{2} – c^{2} = 0 ……..(2)

Radical axis of S = 0, S’ = 0 is S – S’ = 0.

⇒ 2bx – 2ax + 2ay – 2by = 0

⇒ 2x(b – a) + 2y(a – b) = 0

⇒ x – y = 0 (∵ a ≠ b)

II.

Question 1.

Find the equation of the common tangent of the following circles at their point of contact.

(i) x^{2} + y^{2} + 10x – 2y + 22 = 0, x^{2} + y^{2} + 2x – 8y + 8 = 0

Solution:

Denote S ≡ x^{2} + y^{2} + 10x – 2y + 22 = 0 and S’ ≡ x^{2} + y^{2} + 2x – 8y + 8 = 0.

The equation of the common tangent of the circles at their point of contact is the equation of the radical axis of S = 0 and S’ = 0 given by S – S’ = 0.

⇒ 8x + 6y + 14 = 0

⇒ 4x + 3y + 7 = 0

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

Solution:

Let S ≡ x^{2} + y^{2} – 8y – 4 = 0 and S’ ≡ x^{2} + y^{2} – 2x – 4y = 0

The equation of the common tangent at the point of contact of circles is the equation of the radical axis of the circles S = 0, S’ = 0 given by S – S’ = 0.

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

⇒ 2x – 4y – 4 = 0

⇒ x – 2y – 2 = 0

Question 2.

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.

Solution:

Let the equations of the given circles be denoted by S ≡ x^{2} + y^{2} – 8x – 2y + 8 = 0 and S’ = x^{2} + y^{2} – 2x + 6y + 6 = 0

Centres are C_{1} = (4, 1) and C_{2} = (1, -3)

and radii of circles are r_{1} = \(\sqrt{16+1-8}\) = 3 and r_{2} = \(\sqrt{1+9-6}\) = 2

∴ C_{1}C_{2} = \(\sqrt{(4-1)^2+(1+3)^2}=\sqrt{9+16}\) = 5

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

Since C_{1}C_{2} = r_{1} + r_{2}, the circles touch each other externally.

The point of contact P divides C_{1}C_{2} internally in the ratio of their radii i.e., 3 : 2.

∴ The point of contact

Question 3.

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 fg’ = f’g.

Solution:

Let S = x^{2} + y^{2} + 2gx + 2fy = 0 and S’ = x^{2} + y^{2} + 2g’x + 2f’y = 0 be the given circles.

Centres of the circles are C_{1} = (-g, -f) and C_{2} = (-g, -f’)

Also r_{1} = \(\sqrt{g^2+f^2}\) and r_{2} = \(\sqrt{g^{\prime 2}+f^{\prime 2}}\)

Given that the two circles touch each other we have C_{1}C_{2} = |r_{1} ± r_{2}|

⇒ \(\sqrt{\left(g-g^{\prime}\right)^2+\left(f-f^{\prime}\right)^2}\) = \(\left|\sqrt{g^2+f^2} \pm \sqrt{g^{+2}+f^2}\right|\)

Question 4.

Find the radical centre of the following circles.

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

Solution:

Let S = x^{2} + y^{2} – 4x – 6y + 5 = 0, S_{1} = x^{2} + y^{2} – 2x – 4y – 1 = 0 and S_{2} = x^{2} + y^{2} – 6x – 2y = 0 be the given circles.

Radical axis of S = 0, S_{1} = 0 is S – S_{1} = 0

⇒ -2x – 2y + 6 = 0

⇒ x + y – 3 = 0 ……….(1)

Radical axis of S_{1} = 0, S_{2} = 0 is 4x – 2y – 1 = 0 ……….(2)

Solving (1) and (2), we get the coordinates 0f Radical centre of the circles S = 0, S_{1} = 0, and S_{2} = 0

(ii) x^{2} + y^{2} + 4x – 7 = 0, 2x^{2} + 2y^{2} + 3x + 5y – 9 = 0, x^{2} + y^{2} + y = 0

Solution:

Denote S = x^{2} + y^{2} + 4x – 7 = 0

S_{1} = \(x^2+y^2+\frac{3}{2} x+\frac{5}{2} y-\frac{9}{2}=0\)

and S_{2} = x^{2} + y^{2} + y = 0

Radical axis of S = 0, S_{1} = 0 is S – S_{1} = 0.

⇒ \(\frac{5}{2} x-\frac{5}{2} y+\frac{5}{2}=0\)

⇒ x – y + 1 = 0 ………(1)

Radical axis of S_{1} = 0, S_{2} = 0 is S_{1} – S_{2} = 0.

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

⇒ x + y – 3 = 0 ……..(2)

Solving 2x – 2 = 0 ⇒ x = 1

and From (2), y = 2

∴ Radical centre = (1, 2)

III.

Question 1.

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

Solution:

Let S = x^{2} + y^{2} – 6x – 4y + 9 = 0 and S’ = x^{2} + y^{2} – 8x – 6y + 23 = 0

Then the equation of the common chord of circles is S – S’ = 0 is 2x + 2y – 14 = 0

⇒ x + y – 7 = 0

The Centre of S’ = 0 is (4, 3)

∴ x + y – 7 = 4 + 3 – 7 = 0 and common chord of circles S = 0, S’ = 0 is the diameter of the second circle.

The length of the common chord = 2(the radius of the circle S’ = 0) = 2√2 units.

(or) C_{1}P = ⊥ distance from C_{1}(3, 2) to the common chord x + y – 7 = 0.

∴ C_{1}P = \(\left|\frac{3+2-7}{\sqrt{2}}\right|\) = √2

AC_{1} = radius of S = 0

= \(\sqrt{9+4-9}\)

= 2

∴ AP = \(\sqrt{\mathrm{AC}_1^2-\mathrm{C}_1 \mathrm{P}^2}=\sqrt{4-2}=\sqrt{2}\)

∴ Length of the common chord AB = 2(AP) = 2√2.

Question 2.

Find the equation and length of the common chord of the following circles.

(i) x^{2} + y^{2} + 2x + 2y + 1 = 0, x^{2} + y^{2} + 4x + 3y + 2 = 0

Solution:

Let S = x^{2} + y^{2} + 2x + 2y + 1 = 0 and S_{1} = x^{2} + y^{2} + 4x + 3y + 2 = 0

Then the equation of the common chord of S = 0, S_{1} = 0 is S – S_{1} = 0.

⇒ -2x – y – 1 = 0

⇒ 2x + y + 1 = 0

C_{1} = Centre of S = 0 is (-1, 1) and r_{1} = \(\sqrt{1+1-1}\) = 1

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

Solution:

Denote S = x^{2} + y^{2} – 5x – 6y + 4 = 0 and S_{1} = x^{2} + y^{2} – 2x – 2 = 0

Equation of common chord of S = 0, S_{1} = 0 is S – S_{2} = 0.

⇒ -3x – 6y + 6 = 0

⇒ x + 2y – 2 = 0

Question 3.

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 bisector the circumference of the latter) if 2g'(g – g’) + 2f(f – f’) = c – c’.

Solution:

Let S ≡ x^{2} + y^{2} + 2gx + 2fy + c = 0 and S’ ≡ x^{2} + y^{2} + 2g’x + 2f’y + c’ = 0 be the given circles.

Centre of the circle S’ = 0 is C2 = (-g’, -f’)

The equation of the Radical axis of S = 0 and S’ = 0 is S – S’ = 0.

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

Substituting the centre (-g’, -f’) of S’ = 0, we get

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

⇒ 2g'(g – g’) + 2f'(f – f’) = c – c’

The radical axis of the circles S = 0, S’ = 0 is the diameter of the circle S’ = 0 if 2g'(g – g’) + 2f(f – f’) = c – c’

Question 4.

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:

Let S = x^{2} + y^{2} + 2ax + c = 0 and S’ = x^{2} + y^{2} + 2by + c = 0 be the given circles.

Centre of S = 0 is C_{1} (-a, 0) and radius of the circle S = 0 is \(\sqrt{a^2-c}\).

The equation of common tangent of circles S = 0, S’ = 0 is S – S’ = 0

⇒ 2ax – 2by = 0

⇒ ax – by = 0

Let L = ax – by = 0

The two circles S = 0 and S’ = 0 touch each other if the perpendicular distance from C_{1} (-a, 0) to the common tangent L = 0 is equal to the radius of the circle S = 0.

∴ The perpendicular distance from C_{1} (-a, 0)

Question 5.

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:

Let S = x^{2} + y^{2} – 2x = 0 and S’ = x^{2} + y^{2} + 6x – 6y + 2 = 0 be the given circles.

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

and radius of S_{1} = 0 is r_{1} = √1 = 1.

Centre of S’ = 0 is C_{2} = (-3, 3)

and radius of S’ = 0 is r_{2} = \(\sqrt{9+9-2}\) = 4.

Distance between centres C_{1}C_{2} = \(\sqrt{(1+3)^2+(0-3)^2}=\sqrt{16+9}\) = 5

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

Since C_{1}C_{2} = r_{1} + r_{2}, the two circles touch externally at a point and the point of contact divides C_{1}C_{2} internally in the ratio 1 : 4.

∴ Point of contact

∴ The point of contact is \(\left(\frac{1}{5}, \frac{3}{5}\right)\) and the two circles touch each other externally.

Question 6.

Find the equation of the circle which cuts the following circles orthogonally.

(i) x^{2} + y^{2} + 4x – 7 = 0, 2x^{2} + 2y^{2} + 3x + 5y – 9 = 0, x^{2} + y^{2} + y = 0

Solution:

Let S = x^{2} + y^{2} + 4x – 7 = 0

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

and S” = x^{2} + y^{2} + y = 0 be the equations of the given circles.

Radical axis of S = 0, S’ = 0 is S – S’ = 0.

⇒ 5x – 5y – 5 = 0

⇒ x – y – 1 = 0 ………(1)

The radical axis of S’ = 0, S” = 0 is S’ – S” = 0.

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

⇒ x + y – 3 = 0 ………(2)

Solving (1) and (2), we get

the Radical centre of circles S = 0, S’ = 0 and S” = 0

⇒ x = 2 and y = 1

∴ The radical centre of the given circles is (2, 1).

The length of the tangent drawn from (2, 1) to S = 0 is \(\sqrt{4+1+8-7}\) = √6

So by the property that a circle having radical centre as centre and length of the tangent to S = 0 as radius cuts every given circle orthogonally.

So the equation of a required circle with C(2, 1) as a radical centre and √6 as the radius is given by

(x – 2)^{2} + (y + 1)^{2} = (√6)^{2}

⇒ x^{2} + y^{2} – 4x – 2y – 1 = 0

(ii) x^{2} + y^{2} + 2x + 4y + 1 = 0, 2x^{2} + 2y^{2} + 6x + 8y – 3 = 0, x^{2} + y^{2} – 2x + 6y – 3 = 0. (May ’12)

Solution:

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

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

and S” = x^{2} + y^{2} – 2x + 6y – 3 = 0 be the equations of the given circles.

Radical axis of centre S = 0, S’ = 0 is S – S’ = 0.

⇒ -x + \(\frac{5}{2}\) = 0

⇒ x = \(\frac{5}{2}\) ……..(1)

Radical axis of centre S’ = 0, S” = 0 is S’ – S” = 0

⇒ 5x – 2y + \(\frac{3}{2}\) = 0

⇒ 10x – 4y + 3 = 0 ………(2)

By solving (1) and (2), we get the Radical centre of the given circles.

Since x = \(\frac{5}{2}\) and from(2),

25 – 4y + 3 = 0

⇒ y = 7

∴ Radical centre = (\(\frac{5}{2}\), 7)

Length of the tangent from (\(\frac{5}{2}\), 7) to S = 0 is

Hence the circle with centre (\(\frac{5}{2}\), 7) and \(\sqrt{\frac{357}{4}}\) as radius cuts every circle of the given system orthogonally.

∴ The equation of required circle with centre (\(\frac{5}{2}\), 7) and \(\sqrt{\frac{357}{4}}\) as radius is

(iii) x^{2} + y^{2} + 2x + 17y + 4 = 0, x^{2} + y^{2} + 7x + 6y + 11 = 0, x^{2} + y^{2} – x + 22y + 3 = 0

Solution:

Denote the given circles by

S = x^{2} + y^{2} + 2x + 17y + 4 = 0

S’ = x^{2} + y^{2} + 7x + 6y + 11 = 0

S” = x^{2} + y^{2} – x + 22y + 3 = 0

Radical axis of S = 0, S’ = 0 is S – S’ = 0

⇒ -5x + 11y – 7 = 0

⇒ 5x – 11y + 7 = 0 ………(1)

Radical axis of S’ = 0, S” = 0 is S’ – S” = 0

⇒ 8x – 16y + 8 = 0

⇒ x – 2y + 1 = 0 ……..(2)

Solving (1) and (2), we get the coordinates of the Radical centre of circles.

∴ \(\frac{x}{-11+14}=\frac{y}{7-5}=\frac{1}{-10+11}\)

⇒ x = 3, y = 2

∴ Radical centre = (3, 2)

Length of the tangent from (3, 2) to the circle S = 0 is \(\sqrt{9+4+6+34+4}=\sqrt{57}\).

Since the circle with centre (3, 2) and radius √57 cuts every circle of the given system orthogonally,

the equation of the required circle is (x – 3)^{2} + (y – 2)^{2} = 57

⇒ x^{2} + y^{2} – 6x – 4y – 44 = 0.

(iv) 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:

Denote the given circles by S = x^{2} + y^{2} + 4x + 2y + 1 = 0, S’ = x^{2} + y^{2} + 4x + 3y – \(\frac{3}{2}\) = 0, and S” = x^{2} + y^{2} + 6x – 2y – 3 = 0

Radical axis of S = 0, S’ = 0 is S – S’ = 0

⇒ -y + \(\frac{5}{2}\) = 0

⇒ y = \(\frac{5}{2}\) ……..(1)

The radical axis of S’ = 0, S” = 0 is S’ – S” = 0.

⇒ -2x + 5y + \(\frac{3}{2}\) = 0

⇒ -4x + 10y + 3 = 0

⇒ 4x – 10y – 3 = 0 ………(2)

Solving (1) and (2), we get the Radical centre of circles.

∴ From (2), 4x – 25 – 3 = 0

⇒ x = 7

∴ Radical centre = (7, \(\frac{5}{2}\))

Length of the tangent from (7, \(\frac{5}{2}\)) to S = 0 is = \(\sqrt{49+\frac{25}{4}+28+5+1}=\sqrt{\frac{357}{4}}\)

∴ The equation of the required circle with centre (7, \(\frac{5}{2}\)) and as radius which cuts every circle of the system orthogonally is given by

(x – 7)^{2} + \(\left(y-\frac{5}{2}\right)^2=\frac{357}{4}\)

⇒ x^{2} + y^{2} – 14x – 5y + 49 + \(\frac{25}{4}-\frac{357}{4}\) = 0

⇒ x^{2} + y^{2} – 14x – 5y – 34 = 0.