Question

Two circles are inscribed and circumscribed about a square $ABCD$, length of each side of the square is $32.P$ and $Q$ are two points respectively on these circles, then $\sum (PA{)}^2+\sum (QA{)}^2$ is equal to

Answer 1

Let centre of the square be the origin $O$ and the lines through $O$ parallel to the sides of the square are the coordinate axes.

Then the vertices of the square are $A(16,16),B(-16,16),C(-16,-16)$ and $D(16,-16)$

The radii of the inscribed and circumscribed circles are respectively $16$ and $OA=\sqrt{{16}^2+{16}^2}=16\sqrt{2}$

and their centre is at the origin.

Let the coordinate of $P$ be $(16\cos \theta ,16\sin \theta )$ and that of $Q$ be $(16\sqrt{2}\cos \varphi ,16\sqrt{2}\sin \varphi )$.

Then, $\sum (PA{)}^2={16}^2[(\cos \theta -1{)}^2+(\sin \theta -1{)}^2+(\cos \theta +1{)}^2+(\sin \theta -1{)}^2+(\cos \theta +1{)}^2+$

$(\sin \theta +1{)}^2+(\cos \theta -1{)}^2+(\sin \theta +1{)}^2]$

$=2\times {16}^2[2{\cos }^2\theta +2+2{\sin }^2\theta +2]=12\times (16{)}^2$

Similarly $\sum (QA{)}^2=16\times (16{)}^2$

$\sum (PA{)}^2+\sum (QA{)}^2=28\times (16{)}^2=7168$