The equation of the circle which passes through the origin and cuts orthogonally each of the circles…
The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6 x+8=0$ and $x^2+y^2-2 x-2 y=7$ is
$3 x^2+3 y^2-8 x-13 y=0$
$3 x^2+3 y^2+8 x+29 y=0$
$3 x^2+3 y^2+8 x+29 y=0$
$3 x^2+3 y^2-8 x-29 y=0$
Solution
Let the required equation of circle be $x^2+y^2+2 g x+2 f y=0$. Since, the above circle cuts the given circles orthogonally.
$\begin{aligned}
& \therefore \quad 2(-3 g)+2 f(0)=8 \\
& \Rightarrow \\
& 2 g=-\frac{8}{3} \\
& \text { and } \\
& -2 g-2 f=-7 \\
& \Rightarrow \\
& 2 f=+7+\frac{8}{3}=\frac{29}{3} \\
&
\end{aligned}$
$\therefore$ Required equation of circle is
or
$\begin{aligned}
& x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0 \\
& 3 x^2+3 y^2-8 x+29 y=0
\end{aligned}$