Question

The equation of the circle whose diameter is the common chord of the circles

x²+ y² + 3x + 2y + 1 = 0 and x² + y² + 3x + 4y + 2 = 0 is

• A. x² + y² + 8x + 10y + 2 = 0
• B. x²+ y² – 5x + 4y + 7 = 0
• C. 2x² + 2y² + 6x + 2y + 1 = 0
• D. None of these

Answer 1

Answer: (C)

2x² + 2y² + 6x + 2y + 1 = 0

Answer 2

The equation of the common chord of the circles

x² + y² + 3x + 2y + 1 = 0 and

x² + y² + 3x + 4y + 2 = 0 is 2y + 1 = 0. (Using S₁-S₂ = 0)

The equation of a circle passing through the intersection of

x²+y²+3x+2y+1 = 0 and x²+y²+3x+4y+2 = 0 is

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

⇒ x²+y²+x$\left( \frac{3 + 3\lambda}{\lambda + 1} \right) + 2y\mspace{6mu}\left( \frac{1 + 2\lambda}{\lambda + 1} \right)\mspace{6mu} + \frac{1 + 2\lambda}{\lambda + 1} = 0$.… (1)

Since 2y + 1 = 0 is a diameter of this circle, therefore its centre $\left( - \frac{3 + 3\lambda}{2},\mspace{6mu} - \frac{2\lambda + 1}{\lambda + 1} \right)$ lies on 2y + 1 = 0 i.e.

-2 $\left( \frac{2\lambda + 1}{\lambda + 1} \right)\mspace{6mu} + 1\mspace{6mu} = \mspace{6mu} 0$

⇒ -4λ - 2 + λ +1 = 0 ⇒ - 3λ = 1 ⇒ λ = –$\frac{1}{3}$.

Putting λ = –$\frac{1}{3}$ in (1), the equation of the required circle is 2x²+2y²+6x+2y+1 = 0