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

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

I.

Question 1.

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

i) x² + y² – 3x – 4y + 5 = 0, 3(x² + y²) – 7x + 8y – 11 = 0

Solution:

S ≡ x² + y² – 3x – 4y + 5 = 0

S ≡ 3x² + 3y² – 7x + 8y + 11 = 0

S – S’ = 0 is radical axis.

(x² + y² – 3x – 4y + 5)

ii) x² + y² + 2x + 4y + 1 = 0, x² + y² + 4x + y = 0.

Solution:

S – S’ = 0 is radical axis.

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

⇒ -2x + 3y + 1 = 0

(or) 2x – 3y – 1 =0 required radical axis.

iii) x² + y² +4x + 6y – 7 = 0,

4(x² + y²) + 8x + 12y – 9 = 0.

Solution:

S – S’ = 0 is radical axis.

(x² + y² + 4x + 6y – 7) – (x² + y² + 2x + 3y – \(\frac{9}{4}\)) = o

⇒ 2x + 3y – \(\frac{19}{4}\) = 0 ⇒ 8x + 12y – 19 = 0

iv) x² + y² – 2x – 4y -1=0, x² + y² – 4x – 6y + 5 = 0.

Solution:

S – S’ = 0 radical axis

(x² + y² – 2x – 4y – 1) – (x² + y² – 4x – 6y + 5) = 0

2x + 2y – 6 = 0 (or) x + y – 3 = 0

Question 2.

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

i) x² + y² – 4x – 4y + 3 = 0, x² + y² – 5x – 6y + 4 = 0.

Solution:

(x² + y² – 4x – 4y + 3) – (x² + y² – 5x – 6y + 4) = 0

x + 2y – 1 = 0 Equation of common chord.

ii) x² + y² + 2x + 3y + 1 = 0, x² + y² + 4x + 3y + 2 = 0.

Solution:

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

-2x – 1 = 0 equation of common chord is

iii) (x – a)² + (y – b)² = c², (x – b)² + (y – a)² = c² (a ≠ b)

Solution:

(x² + y² – 2xa – 2yb – c²) – (x² + y² – 2xb – 2ya – c²) = 0

-2x(a – b) – 2y(b – a) = 0

(or) x – y = 0

II.

Question 1.

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

i) x² + y² + 10x – 2y + 22 = 0, x² + y² + 2x – 8y + 8 = 0.

Solution:

x² + y² + 10x – 2y + 22 = 0

x² + y² + 2x – 8y + 8 = 0

When circles touch each other then

S – S’ = 0 is required tangent (common)

∴ (x² + y² + 10x-2y + 22) – (x² + y² + 2x – 8y + 8) = 0

8x + 6y + 14 = 0 (or)

4x + 3y + 7 = 0

ii) x² + y² – 8y – 4 = 0; x² + y² – 2x – 4y = 0.

Solution:

When circles touch each other then

S – S’ = 0 is required common tangent.

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

2x – 4y – 4 = 0 (or) x – 2y – 2 = 0

Question 2.

Show that the circles x² + y² – 8x – 2y + 8 = 0 and x² + y² – 2x +, 6y + 6 = 0 touch each other and find the point of contact.

Solution:

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

r_{1} = \(\sqrt{16+1-8}\) = 3 ; r_{2} = \(\sqrt{1+9-6}\) = 2

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

r_{1} + r_{2} = C_{1} + C_{2} they touch each other externally

∴ Point of contact is (\(\frac{11}{5} , \frac{-7}{5}\))

Question 3.

If the two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’.

Solution:

Question 4.

Find the radical centre of the following circles.

i) x² + y² – 4x – 6y + 5 = 0 ………… (i)

x² + y² – 2x – 4y – 1 = 0 ………… (ii)

x² + y² – 6x – 2y = 0 ………… (iii)

Solution:

(i) – (ii) gives

– 2x – 2y + 6 = 0

x + y – 3 = 0 ………. (1)

(ii) – (iii) gives

Point of intersection of (1) and (2) is radical centre will (7/6,11/6) we get by solving these two equations.

ii) x² + y² + 4x – 7 = 0,

2x² + 2y² + 3x + 5y – 9 = 0,

x² + y² + y = 0.

Solution:

S = x² + y² + 4x – 7 = 0 ……….. (i)

S_{1} = 2x² + 2y² + 3x + 5y – 9 = 0

= x² + y² + \(\frac{3}{2}\)x + \(\frac{5}{2}\)y – \(\frac{9}{2}\) = 0 ………. (ii)

S_{11} = x² + y² + y = 0 ………. (iii)

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

5x – 5y – 5 = 0

x – y – 1 = 0 ………… (iv)

Radical axis of S = 0, S_{11} = 0 is S – S_{11} = 0

4x – y – 7 = 0 ……… (v)

x – y – 1 = 0………….. (vi)

Subtracting 3x – 6 = 0 ⇒ 3x = 6

x = \(\frac{6}{3}\) = 2

Substituting in (iv), 2 – y – 1 = 0

y = 1

Radical centre is P(2, 1)

III.

Question 1.

Show that the common chord of the circles x² + y² – 6x – 4y + 9 = 0 and x² + y² – 8x – 6y + 23 = 0 is the diameter of the second circle and also find its length.

Solution:

Common chord be

(x² + y² – 6x – 4y + 9) – (x² + y² – 8x – 6y + 23) = 0

2x + 2y – 14 = 0

x + y – 7 = 0 ………… (i)

Centre of circle (-4, -3)

(-4, -3) lies on line x + y – 7

Radius is {4² + 3² – 23}^{½} = √2

Diameter = 2√2

Question 2.

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

i) x² + y² + 2x + 2y + 1 =0,

x² + y² + 4x + 3y + 2 = 0.

Solution:

x² + y² + 2x + 2y + 1 = 0

x² + y² + 4x + 3y + 2 = 0

Equation of common chord is

S – S’ = 0 (x² + y² + 2x + 2y + 1) – (x² + y² + 4x + 3y + 2) = 0

-2x – y – 1 = 0

2x + y + 1 = 0

Centre of circle is (-1, -1)

Radius = \(\sqrt{1+1-1}\) = 1

Length of ⊥ from centre (-1, -1) to the chord is

d = \(|\frac{2(-1)+(-1)+1}{\sqrt{2^{2}+1^{2}}}|=\frac{2}{\sqrt{5}}\)

ii) x² + y² – 5x – 6y + 4 = 0, x² + y² – 2x – 2 = 0

Solution:

Common chord equation

(x² + y² – 5x- 6y + 4) – (x²+ y² – 2x – 2) = 0

-3x – 6y + 6 = 0

x + 2y – 2 = 0

Question 3.

Prove that the radical axis of the circles x² + y² + 2gx + 2fy + c = 0 and x² +y² + 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:

Radical axis is

(x² + y² + 2gx + 2fy + c) – (x² + y² + 2g’x + 2fy + c’) = 0

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

Centre of second circle is (-g1, -f1)

Radius = \(\sqrt{g’^{2}+f’^{2}+c’}\)

Now (-g’, -f’) should lie on (i)

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

(or) 2g (g – g’) + 2f'(f – f’) = c – c’

Question 4.

Show that the circles x² + y² + 2ax + c = 0 and x² + y² + 2by + c = 0 touch each other if 1/a² + 1/b² = 1/c.

Solution:

The centres of the circles C_{1} (-0, 0) and C_{2} (0, -b) respectively

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

Radius of 2nd circle be \(\sqrt{b^{2}-c}\) = r_{2}

C_{1}C_{2} = r_{1} + r_{2}

(C_{1} C_{2})² = (r_{1} + r_{2})²

(a² + b²) = a² – c + b² – c + 2\(\sqrt{a^{2}-c}.\sqrt{a^{2}-c}\)

c = \(\sqrt{a^{2}-c}.\sqrt{a^{2}-c}\)

c² = (a² – c) (b² – c)

c² = -c (a² + b²) + a²b² + c²

(or) c(a² + b²) = a²b² (or) \(\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)

Question 5.

Show that the circles x² + y² – 2x = 0 and x² + y² + 6x – 6y + 2 = 0 touch each other. Find the coordinates of the point of contact. Is the point of contact external or internal?

Solution:

For the circle S = x² + y² – 2x = 0

Centre C_{1} = (1, 0) and Radius r_{1} .= \(\sqrt{1+0}\) = 1

For the circle S’ = x² + y² + 6x – 6y + 2 = 0

Centre C_{2} = (-3, 3) and

As C_{1}C_{2} = r_{1} + r_{2} the two circles touch each other externally, the point of contact P divides line of centres internally in the ratio

r_{1} : r_{2} = 1 : 4

Hence point of contact

The contact of the circle is external.

Question 6.

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

i) x² + y² + 4x – 7 = 0,

2x² + 2y² + 3x + 5y – 9 = 0,

x² + y² + y = 0.

Solution:

S = x² + y² + 4x – 7 = 0 ………… (i)

S_{1} = 2x² + 2y² + 3x + 5y – 9 = 0

x – y – 1 = 0 ……. (iv)

Radical axis of S = 0, S_{11} = 0 is S – S_{11} = 0

4x – y – 7 = 0 ……….. (v)

Solving (iv) and (v)

We get 3x – 6 = 0

x = -2

Substute x value in (iv), 2 – y – 1 =0

y = 1

Radical centre is P(2, 1)

PT =’ Length of the tangent from P to S = 0

= \(\sqrt{4+1+8-7}\)

= √6

Equation of the circles cutting the given circles orthogonally

(x – 2)² + (y – 1)² = (√6)²

x² + 4 – 4x + y² + 1 – 2y = 6

x² + y² – 4x – 2y – 1 = 0

ii) x² + y² + 2x + 4y + 1 = 0,

2x² + 2y² + 6x + 8y – 3 = 0,

x² + y² – 2x + 6y – 3 = 0.

Solution:

Equations of the required circles are

S ≡ x² + y² + 2x + 4y + 1 = 0

S_{1} ≡ x² + y² + 3x + 4y – \(\frac{3}{2}\) = 0

S_{11} ≡ x² + y² – 2x + 6y – 3 = 0

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

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

Radical axis of S = 0, S_{11} = 0 is S – S_{11} = 0

4x – 2y + 4 = 0

⇒ 2x – y + 2 = 0

x = \(\frac{5}{2}\) ⇒ 5 – y + 2 = 0

⇒ y = 7

Radical centre is P (\(\frac{5}{2}\) , 7)

PT = Length of the tangent from P to S = 0

Equation of the circles cutting the given circles orthogonally

iii) x² + y² + 2x + 17y + 4 = 0,

x² + y² + 7x + 6y + 11 = 0,

x² + y² – x + 22y + 3 = 0

Solution:

Equations of the required circles are

S ≡ x² + y² + 2x + 17y + 4 = 0 ……… (i)

S_{1} ≡ x² + y² + 7x + 6y + 11 = 0 ……… (ii)

S_{11} ≡ x² + y² – x + 22y + 3 = 0 ……… (iii)

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

-5x + 11 y – 7 = 0

5x – 11y + 7 = 0 ……… (iv)

Radical axis of S = 0, S_{11} = 0 is S – S_{11} =0

3x – 5y + 1 = 0 ………… (v)

Solving (iv) and (v)

Radical centre is P(3, 2)

PT = Length of the tangeqt from P to S = 0

= \(\sqrt{9+4+6+34+4}\) = √57

Equation of the circle cutting the given circles cutting orthogonally

(x – 3)² + (y – 2)² = 57

x² – 6x + 9 + y² – 4y + 4 – 57 = 0

x² + y² – 6x – 4y – 44 = 0

iv) x² + y² + 4x + 2y + 1 = 0,

2(x² + y²) + 8x + 6y – 3 = 0,

x² + y² + 6x – 2y – 3 = 0.

Solution:

Equations of the required circles are

S ≡ x² + y² + 4x + 2y + 1 = 0 ……….. (i)

S_{1} ≡ x² + y² + 4x + 3y – \(\frac{3}{2}\) = 0 ………… (ii)

S_{11} ≡ x² + y² + 6x – 2y – 3 = 0 ………. (iii)

(i) – (ii) gives radical axis of S = 0, S_{1} = 0 is

S – S_{1} = 0 ⇒ -y + \(\frac{5}{2}\) = 0 ⇒ y = \(\frac{5}{2}\)

Radical axis of S = 0, S_{11} = 0 is S – S_{11} = 0

– 2x + 4y + 4 = 0

x – 2y – 2 = 0

y = \(\frac{5}{2}\) ⇒ x – 5 – 2 = 0

x = 5 + 2 = 7

Radical centre is P (7, \(\frac{5}{2}\))

PT = Length of the tangent P to S = 0

Equation of the required circle is