The ends of the hypotenuse of a right angled triangle are (0, 6) and (6, 0). The locus of its third vertex is

Given A (6, 0), B (0, 6) are ends of hypotenuse of right angled triangle.

Let P (x, y) be locus of third vertex 

since $\angle \mathrm{APB}={90}^0$ then

$$\begin{gathered} {\mathrm{PA}}^2+{\mathrm{PB}}^2={\mathrm{AB}}^2 \\ \Rightarrow {(\mathrm{x}-6)}^2+{\mathrm{y}}^2+{\left(\mathrm{x}\right)}^2+{(\mathrm{y}-6)}^2={\left(6\right)}^2+{\left(6\right)}^2 \Rightarrow {\mathrm{x}}^2+36-12\mathrm{x}+{\mathrm{y}}^2+{\mathrm{x}}^2+{\mathrm{y}}^2+36-12\mathrm{y}=72 \\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2-12\mathrm{x}-12\mathrm{y}=0 \\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}-6\mathrm{y}=0 \end{gathered}$$