Find the equation of the circle which cuts orthogonally the circle x²+y²–4x+2y–7 = 0 and having centre at (2,3).

The given circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}+2\mathrm{y}-7=0$ ...(1)
Let the required circle which cuts orthogonally the circle (i) is
${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 ...\left(2\right)$
given its centre is (–g, –f) = (2, 3) $\Rightarrow$g = –2,f = –3
Since  two circles (1) and (2) are cutting each other orthogonally.

then $2{\mathrm{g}}_1{\mathrm{g}}_2+2{\mathrm{f}}_1{\mathrm{f}}_2={\mathrm{c}}_1+{\mathrm{c}}_2$
$\begin{array}{l}2(-2)(-2)+2(-3)\left(1\right)=-7+\mathrm{c}\Rightarrow 8-6=-7+\mathrm{c}\\ \Rightarrow +2=-7+\mathrm{c}\Rightarrow \mathrm{c}=7+2=9\Rightarrow \mathrm{c}=9\end{array}$
Hence the required circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-6\mathrm{y}+9=0$