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:
C1 = (4, 1) C2 = (1, -3)
r1 = $$\sqrt{16+1-8}$$ = 3 ; r2 = $$\sqrt{1+9-6}$$ = 2
C1C2 = $$\sqrt{(4-1)^{2}+(1+3^{2})}$$ = 5
r1 + r2 = C1 + C2 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)
S1 = 2x² + 2y² + 3x + 5y – 9 = 0
= x² + y² + $$\frac{3}{2}$$x + $$\frac{5}{2}$$y – $$\frac{9}{2}$$ = 0 ………. (ii)
S11 = x² + y² + y = 0 ………. (iii)
Radical axis of S = 0, S1 = 0 is S – S1 = 0

5x – 5y – 5 = 0
x – y – 1 = 0 ………… (iv)
Radical axis of S = 0, S11 = 0 is S – S11 = 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:
(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 C1 (-0, 0) and C2 (0, -b) respectively
Radius of 1st circle be $$\sqrt{a^{2}-c}$$ = r1
Radius of 2nd circle be $$\sqrt{b^{2}-c}$$ = r2
C1C2 = r1 + r2
(C1 C2)² = (r1 + r2
(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 C1 = (1, 0) and Radius r1 .= $$\sqrt{1+0}$$ = 1
For the circle S’ = x² + y² + 6x – 6y + 2 = 0
Centre C2 = (-3, 3) and

As C1C2 = r1 + r2 the two circles touch each other externally, the point of contact P divides line of centres internally in the ratio
r1 : r2 = 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)
S1 = 2x² + 2y² + 3x + 5y – 9 = 0

x – y – 1 = 0 ……. (iv)
Radical axis of S = 0, S11 = 0 is S – S11 = 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
S1 ≡ x² + y² + 3x + 4y – $$\frac{3}{2}$$ = 0
S11 ≡ x² + y² – 2x + 6y – 3 = 0
Radical axis of S = 0, S1 = 0 is S – S1 =0
-x + $$\frac{5}{2}$$ = 0 ⇒ x = $$\frac{5}{2}$$
Radical axis of S = 0, S11 = 0 is S – S11 = 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)
S1 ≡ x² + y² + 7x + 6y + 11 = 0 ……… (ii)
S11 ≡ x² + y² – x + 22y + 3 = 0 ……… (iii)
Radical axis of S = 0, S1 = 0 is S – S1 = 0
-5x + 11 y – 7 = 0
5x – 11y + 7 = 0 ……… (iv)
Radical axis of S = 0, S11 = 0 is S – S11 =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)
S1 ≡ x² + y² + 4x + 3y – $$\frac{3}{2}$$ = 0 ………… (ii)
S11 ≡ x² + y² + 6x – 2y – 3 = 0 ………. (iii)
(i) – (ii) gives radical axis of S = 0, S1 = 0 is
S – S1 = 0 ⇒ -y + $$\frac{5}{2}$$ = 0 ⇒ y = $$\frac{5}{2}$$
Radical axis of S = 0, S11 = 0 is S – S11 = 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