Find the equation of locus of P, if the line segment joining (2, 3) and (–1, 5) subtends a right angle at P.

Given points are A(2, 3), B(–1, 5)
Let P(x, y) is any point on the locus
$\text{ Given condition is }\mathrm{\angle }\mathrm{APB}={90}^{\circ }$
${\mathrm{AP}}^2+{\mathrm{PB}}^2={\mathrm{AB}}^2$
$\begin{array}{l}\Rightarrow (\mathrm{x}-2{)}^2+(\mathrm{y}-3{)}^2+(\mathrm{x}+1{)}^2+(\mathrm{y}-5{)}^2=(2+1{)}^2+(3-5{)}^2\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4+{\mathrm{y}}^2-6\mathrm{y}+9+{\mathrm{x}}^2+2\mathrm{x}+1+{\mathrm{y}}^2-10\mathrm{y}+25=9+4\\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2-2\mathrm{x}-16\mathrm{y}+26=0\end{array}$
$\text{ locus of }\mathrm{P}(\mathrm{x},\mathrm{y})\text{ is }{\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{x}-8\mathrm{y}+13=0$