Question

Find the equation to the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y=7$.

Answer 1

Let a circle passing through origin be $x^2+y^2+2gx+2fy=\mathrm{0...........}\left(1\right)$

For two circles to be orthogonal, $2g{g}^{\prime }+2f{f}^{\prime }=c+c^{\prime }$

In $x^2+y^2-6x+8=0,g^{\prime }=-3,f^{\prime }=0,c^{\prime }=8$(Comparing with $x^2+y^2+2gx+2fy+c=0)$

In $x^2+y^2-2x-2y-7=0,g^{\prime }=-1,f^{\prime }=-1,c^{\prime }=-7$

So if circle $\left(1\right)$ is orthogonal with $x^2+y^2-6x+8=0$, then

$2.g(-3)=8$

or $g=\frac{-4}{3}$

If circle $\left(1\right)$ is orthogonal with $x^2+y^2-2x-2y-7=0$, then

$2.g(-1)+2f(-1)=-7$

or,$-2\left(\frac{-4}{3}\right)+(-2f)=-7$

or,$-2f=-7-\frac{8}{3}$

or,$-2f=-\frac{29}{3}$

or, $f=\frac{29}{6}$

So, the circle is $x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$

or, $3x^2+3y^2-8x+29y=0$