Find the equation of the circle which is orthogonal to each of the following circles
$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}+2\mathrm{y}+1=0\\ {\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}+6\mathrm{y}+53=0\\ {\mathrm{x}}^2+{\mathrm{y}}^2+5\mathrm{x}-8\mathrm{y}+15=0\end{array}$

$\text{ Given circles are }{\mathrm{S}}^{\mathrm{\prime }}={\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}+2\mathrm{y}+1=0$
$\begin{array}{l}{\mathrm{S}}^{\prime \prime }={\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}+6\mathrm{y}+53=0\\ {\mathrm{S}}^{\prime \prime \mathrm{\prime }}={\mathrm{x}}^2+{\mathrm{y}}^2+5\mathrm{x}-8\mathrm{y}+15=0\end{array}$
Equation of radical axis of $\mathrm{S}=0,{\mathrm{S}}^{\text{'}}=0\text{ is }\mathrm{S}-{\mathrm{S}}^{\text{'}}=0$
$\Rightarrow 4\mathrm{x}-4\mathrm{y}-52=0\Rightarrow \mathrm{x}-\mathrm{y}-13=0 ...\left(1\right)$
$\text{ Equation of radical axis of }{\mathrm{S}}^{\mathrm{\prime }}=0,{\mathrm{S}}^{\prime \prime \mathrm{\prime }}=0\text{ is }$
$\begin{array}{l}{\mathrm{S}}^{\text{'}}-{\mathrm{S}}^{\text{'}\text{'}\text{'}}=0\Rightarrow -2\mathrm{x}+10\mathrm{y}-14=0\\ \Rightarrow \mathrm{x}-5\mathrm{y}+7=0 ...\left(2\right)\end{array}$
$\begin{array}{l}\left(1\right)-\left(2\right)\\ \Rightarrow (-)4\mathrm{y}-20=0\\ \Rightarrow \mathrm{y}=5\end{array}$
Put y = 5 in (2) x = 5y - 7
= 5(5) - 7 = 25 - 7 = 18
$\therefore \text{ radical centre of the given circles is }(18,5)$
Length of the tangent from R.C. (18, 5) to  S' = 0 $\text{ is }\sqrt{{\mathrm{S}}_{11}\text{'}}=\sqrt{{18}^2+5^2+3\left(18\right)+2\left(5\right)+1}$
$=\sqrt{324+25+54+10+1}=\sqrt{414}$
Required circle cuts given circles orthogonally.
$\therefore \text{ Centre of the required circle is R.C. }=(18,5)$
$\text{ Radius of the required circle }\mathrm{r}=\sqrt{{\mathrm{s}}_{"}^{\text{'}}}=\sqrt{414}\text{. }$
$\therefore$Equation of the required circle is ${\left(\mathrm{x}-{\mathrm{x}}_1\right)}^2+{\left(\mathrm{y}-{\mathrm{y}}_1\right)}^2={\mathrm{r}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{x}-18{)}^2+(\mathrm{y}-5{)}^2=414\\ \Rightarrow {\mathrm{x}}^2-36\mathrm{x}+324+{\mathrm{y}}^2-10\mathrm{y}+25-414=0\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-36\mathrm{x}-10\mathrm{y}-65=0.\end{array}$