If point $P$ varies on a circle about the center of rectangle $\square ABCD$, then $PA^2+PB^2+PC^2+PD^2$ remains constant 
Let $\square ABCD$ be a rectangle with center $O$. Prove that if a $P$ is a point that varies over a circle about $O$, then $PA^2+PB^2+PC^2+PD^2$ remains constant

Attempt:


Place $O$ at $(0,0)$ on the Cartesian Plane, and place $P$ at a point $(x,y)$ at a distance $r$ from $O$. Furthermore, place the vertices of the rectangle at the points of the form $(\pm a,\pm b)$, for the corresponding $a$, $b$. Then, $$PA^2+PB^2+PC^2+PD^2=$$ $$2\left(\left(x-a\right)^2+\left(x+a\right)^2+\left(y-b\right)^2+\left(y+b\right)^2\right)=$$ $$4\left(x^2+y^2+a^2+b^2\right)=$$ $$4\left(r^2+a^2+b^2\right),$$ which is clearly a constant. $\blacksquare$
