Find the locus of the third vertex of a right angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4).

Given points A = (4, 0), B = (0, 4)
Let P(x, y) is the third vertex
$\text{ Given condition is }\mathrm{\angle }\mathrm{APB}={90}^{\circ }$
$$\begin{gathered} \Rightarrow {\mathrm{AP}}^2+{\mathrm{PB}}^2={\mathrm{AB}}^2 \\ \Rightarrow (\mathrm{x}-4{)}^2+(\mathrm{y}-0{)}^2+(\mathrm{x}-0{)}^2+(\mathrm{y}-4{)}^2 =(0-4{)}^2+(4-0{)}^2 \\ \begin{array}{l}\Rightarrow {\mathrm{x}}^2-8\mathrm{x}+16+{\mathrm{y}}^2+{\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{y}+16=16+16\\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2-8\mathrm{x}-8\mathrm{y}=0\\ \Rightarrow 2\left({\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}\right)=0\end{array} \\ \therefore \text{ The locus of }\mathrm{P}(\mathrm{x},\mathrm{y})\text{ is }{\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}=0 \end{gathered}$$