Question

The equation of the circle whose diameter is the common chord of the circles
$x^2 + y^2 +2x +3y + 1 = 0 $ and
$x^2 + y^2 + 4x + 3y + 2 = 0 $ is

• A. $2x^2 + 2y^2 + x + 3y + 2 = 0 $
• B. $2x^2 + 2y^2 + 2x + 6y + 1 = 0 $
• C. $2x^2 + 2y^2 + 4x - 3y - 1 = 0 $
• D. $x^2 + y^2 + 2x + 6y - 2 = 0 $

Answer 1

Answer: (B)

The equation of the common chord of the circles
$x^{2}+y^{2}+2 x+3 y+1=0$ and
$x^{2}+y^{2}+4 x+3 y+2=0$ is given by
$2 x+1=0$
[using : $ \left.S_{1}-S_{2}=0\right]$
The equation of a circle passing through the intersection of the given circles is
$\left(x^{2}+y^{2}+2 x+3 y+1\right) $
$+ \lambda\left(x^{2}+y^{2}+4 x+3 y+2\right)=0 $
$\Rightarrow x^{2}(1+\lambda)+y^{2}(1+\lambda)+(1+2 \lambda) $
$ 2 x+3 y(1+\lambda)+1+2 \lambda=0$
$\Rightarrow x^{2}+y^{2}+\left(\frac{1+2 \lambda}{1+\lambda}\right) 2 x+3 y+\frac{1+2 \lambda}{\lambda+1}=0 \ldots$ (i)
Since, $2 x+1=0$ is a diameter of this circle.
Therefore, its centre $\left(-\frac{2 \lambda+1}{\lambda+1},-\frac{3}{2}\right)$ lies on it $\Rightarrow -2\left(\frac{2 \lambda+1}{\lambda+1}\right)+1=0 $
$ \Rightarrow -4 \lambda-2+\lambda+1=0 $
$\Rightarrow -3 \lambda-1=0 $
$ \lambda=-\frac{1}{3}$
On putting $\lambda=-\frac{1}{3}$ in Eq. (i), we get
$\Rightarrow x^{2}+y^{2}+\left(\frac{1-\frac{2}{3}}{1-\frac{1}{3}}\right) 2 x+3 y+\frac{1-\frac{2}{3}}{-\frac{1}{3}+1}=0$
$ \Rightarrow x^{2}+y^{2}+\left(\frac{\frac{1}{3}}{\frac{2}{3}}\right) 2 x+3 y+\frac{\frac{1}{3}}{\frac{3}{3}}=0 $
$ \Rightarrow x^{2}+y^{2}+x+3 y+\frac{1}{2}=0 $
$ \Rightarrow 2 x^{2}+2 y^{2}+2 x+6 y+1=0$