Question

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_{1}=x^{2}+y^{2}+k x-4 y-1=0, S_{2}=3 x^{3}+3 y^{2}-14 x+23 y-15=0$ and passing through the point $(-1,-1)$ is

• A. $x^{2}+y^{2}-8 x-2 y-12=0$
• B. $3 x^{2}+3 y^{2}+18 x-12 y=0$
• C. $5 x^{2}+5 y^{2}-22 x+15 y-17=0$
• D. $x^{2}+y^{2}-5 x+14 y+7=0$

Answer 1

Answer: (C)

We have
$S_{1}: x^{2}+y^{2}+k x-4 y-1=0$
$S_{2}: 3 x^{2}+3 y^{2}-14 x+23 y-15=0 $
$=x^{2}+y^{2}-\frac{14}{3} x+\frac{23}{3} y-5=0$
Since, $S_{1}$ and $S_{2}$ are orthogonal
$\therefore 2\left[\left(\frac{-k}{2}\right)\left(\frac{7}{3}\right)+(2)\left(\frac{-23}{6}\right)\right]=-1-5$
$\Rightarrow 2\left[\frac{-7 k}{6}-\frac{46}{6}\right]=-6$
$ \Rightarrow \frac{-7 k}{6}-\frac{46}{6}=-3 $
$\Rightarrow \frac{7 k}{6}=3-\frac{46}{6}
$$\Rightarrow \frac{7 k}{6}=\frac{18-46}{6}$
$\Rightarrow 7 k=-28 $
$\Rightarrow k=-4 $
$\therefore S_{1}: x^{2}+y^{2}-4 x-4 y-1=0$
$S_{2}: x^{2}+y^{2}-\frac{14}{3} x+\frac{23}{3} y-5=0$
Equation of circle passing through intersection of $S_{1}$ and $S_{2}$ is $x^{2}+y^{2}-4 x-4 y-1+\lambda\left(x^{2}+y^{2}-\frac{14 x}{3}+\frac{23}{3} y-5\right)=0$
Since above circle passes through $(-1,-1)$, so
$1+1+4+4-1+\lambda\left(1+1+\frac{14}{3}-\frac{23}{3}-5\right)=0$
$9-6 \lambda=0$
$ \Rightarrow \lambda=\frac{3}{2}$
$\therefore $ Equation of required circle is
$x^{2}+y^{2}-4 x-4 y-1+\frac{3}{2}\left(x^{2}+y^{2}-\frac{14 x}{3}+\frac{23 y}{3}-5\right)=0 $
$\Rightarrow 2 x^{2}+2 y^{2}-8 x-8 y-2+3 x^{2}+3 y^{2}-14 x+23 y-15=0 $
$\Rightarrow 5 x^{2}+5 y^{2}-22 x+15 y-17=0$