Question

Find the equation to the circle which cuts orthogonally each of the circles $x^2+y^2+2gx+c=0$, $x^2+y^2+2g^{{}^{\prime }}x+c=0$, and $x^2+y^2+2hx+2ky+a=0$.

Answer 1

Let the circle be $x^2+y^2+2px+2qy+d=0$

If two circles are orthogonal then $2g{g}^{\prime }+2f{f}^{\prime }=c+c^{\prime }$

So, from the question

$2.p.g=d+c......\left(1\right)$

$2.p.g^{\prime }=d+c......\left(2\right)$

$2.p.h+2.q.k=d+a......\left(3\right)$

Solving $\left(1\right),\left(2\right)$ and $\left(3\right)$ we get

$p=0$

$q=\frac{a-c}{k}$

$d=-c$

So the circle is

$k{x}^2+k{y}^2+(a-c)y-ck=0$