Question

If a circle passes through the point $(a,b)$ and cuts the circle $x^2+y^2=k^2$ orthogonally, then the locus of its centre is $2ax+2by-(a^2+b^2+k^2)=0$.

Answer 1

Let the circle be $x^2+y^2+2gx+2fy+c=0$.
Since it cuts $x^2+y^2-k^2=0$ orthogonally therefore
$2g\left(0\right)+2f\left(0\right)=c-k^2=0\therefore c=k^2$
Again the circle $C_2$ passes through $(a,b)$
$\therefore a^2+b^2+2ga+2fb+k^2=0\because c=k^2$
$\therefore$ Locus of centre $(-g,-f)$ is
$a^2+b^2+2a(-x)+2b(-y)+k^2=0$
or $2ax+2by-(a^2+b^2+k^2)=0$.