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
$x^2+y^2+2x+3y+1=0$ and
$x^2+y^2+4x+3y+2=0$ is given by
$2x+1=0$
[using : $S_1-S_2=0]$
The equation of a circle passing through the intersection of the given circles is
$\left(x^2+y^2+2x+3y+1\right)$
$+\lambda \left(x^2+y^2+4x+3y+2\right)=0$
$\Rightarrow x^2(1+\lambda )+y^2(1+\lambda )+(1+2\lambda )$
$2x+3y(1+\lambda )+1+2\lambda =0$
$\Rightarrow x^2+y^2+\left(\frac{1+2\lambda }{1+\lambda }\right)2x+3y+\frac{1+2\lambda }{\lambda +1}=0\dots$ (i)
Since, $2x+1=0$ 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
$\Rightarrow x^2+y^2+\left(\frac{1-\frac{2}{3}}{1-\frac{1}{3}}\right)2x+3y+\frac{1-\frac{2}{3}}{-\frac{1}{3}+1}=0$
$\Rightarrow x^2+y^2+\left(\frac{\frac{1}{3}}{\frac{2}{3}}\right)2x+3y+\frac{\frac{1}{3}}{\frac{3}{3}}=0$
$\Rightarrow x^2+y^2+x+3y+\frac{1}{2}=0$
$\Rightarrow 2x^2+2y^2+2x+6y+1=0$