Question

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_1=x^2+y^2+kx-4y-1=0,S_2=3x^3+3y^2-14x+23y-15=0$ and passing through the point $(-1,-1)$ is

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

Answer 1

Answer: (C)

We have
$S_1:x^2+y^2+kx-4y-1=0$
$S_2:3x^2+3y^2-14x+23y-15=0$
$=x^2+y^2-\frac{14}{3}x+\frac{23}{3}y-5=0$
Since, $S_1$ and $S_2$ are orthogonal
$\therefore 2\left[\left(\frac{-k}{2}\right)\left(\frac{7}{3}\right)+(2)\left(\frac{-23}{6}\right)\right]=-1-5$
$\Rightarrow 2\left[\frac{-7k}{6}-\frac{46}{6}\right]=-6$
$\Rightarrow \frac{-7k}{6}-\frac{46}{6}=-3$
$\Rightarrow \frac{7k}{6}=3-\frac{46}{6}<br/>$$\Rightarrow \frac{7k}{6}=\frac{18-46}{6}$
$\Rightarrow 7k=-28$
$\Rightarrow k=-4$
$\therefore S_1:x^2+y^2-4x-4y-1=0$
$S_2:x^2+y^2-\frac{14}{3}x+\frac{23}{3}y-5=0$
Equation of circle passing through intersection of $S_1$ and $S_2$ is $x^2+y^2-4x-4y-1+\lambda \left(x^2+y^2-\frac{14x}{3}+\frac{23}{3}y-5\right)=0$
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
$x^2+y^2-4x-4y-1+\frac{3}{2}\left(x^2+y^2-\frac{14x}{3}+\frac{23y}{3}-5\right)=0$
$\Rightarrow 2x^2+2y^2-8x-8y-2+3x^2+3y^2-14x+23y-15=0$
$\Rightarrow 5x^2+5y^2-22x+15y-17=0$