Find the equation to the locus of the point which is at a distance of 5 units from (–2, 3) in the xy - plane.

Let P(x, y) be a point on the locus A = (–2, 3)
Given PA = 5
S.O.B.S
$\begin{array}{l}{\mathrm{PA}}^2=25\Rightarrow (\mathrm{x}+2{)}^2+(\mathrm{y}-3{)}^2=25\\ {\mathrm{x}}^2+4+4\mathrm{x}+{\mathrm{y}}^2+9-6\mathrm{y}=25\end{array}$
${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}-12=0$
The locus of P(x, y) is
${\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-6\mathrm{y}-12=0$