Question

A circle $C_1$ passes through the origin and has its centre on the line $y=x$. Let $C_1$ cuts $C_2:x^2+y^2-4x-6y+10=0$ orthogonally. If the radius of $C_1$ is $r$, then the value of $8r^2$ is

Answer 1

Let equation of the circle $C_1$ be $x^2+y^2+2gx+2fy+c=0...\left(1\right)$
This passes through $(0,0)$
$\therefore c=0$
The centre $(-g,-f)$ of eqn$\left(1\right)$ lies on the line $y=x$
$\therefore g=f$
Also, eqn$\left(1\right)$ cuts the circle $x^2+y^2-4x-6y+10=0$ orthogonally.
$\therefore 2(-2g-3f)=c+10$
$\Rightarrow -10g=10[\because g=f\text{and}c=0]$
$\Rightarrow g=f=-1$
Hence, equation of the circle $C_1$ is $x^2+y^2-2x-2y=0$
radius, $r=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}$