Two circles are inscribed and circumscribed about a square ABCD, length of each side of the square is 32. $P\text{ and }Q$ are two points respectively on these circles, then $\frac{1}{16}\left[\mathrm{\Sigma }(QA{)}^2-\mathrm{\Sigma }(PA{)}^2\right]$ is equal to

Let the centre of the square be the origin O and the lines through O parallel to the sides of the square be the

coordinates axes. Then the vertices of the square are

$A(16,16),B(-16,16),C(-16,-16)\text{ and }D(16,-16)$

The radii of the inscribed and circumscribed circles are

respectively $16\text{ and }OA=\sqrt{{16}^2+{16}^2}=16\sqrt{2}$ and their centre is at the origin.

Let the coordinate of P be $(16\cos \theta ,16\sin \theta )$ and that of

$\begin{array}{l}Q\text{ be }(16\sqrt{2}\cos \varphi ,16\sqrt{2}\sin \varphi )\\ \text{ then }\mathrm{\Sigma }(PA{)}^2={16}^2\left[(\cos \theta -1{)}^2+(\sin \theta -1{)}^2+\right.\\ (\cos \theta +1{)}^2+(\sin \theta -1{)}^2+(\cos \theta +1{)}^2+(\sin \theta +1{)}^2\\ \left.+(\cos \theta -1{)}^2+(\sin \theta +1{)}^2\right]\\ =2\times {16}^2\left[2{\cos }^2\theta +2+2{\sin }^2\theta +2\right]=12\times (16{)}^2\\ \text{ similarly }\mathrm{\Sigma }(QA{)}^2=16\times (16{)}^2\end{array}$

$\Rightarrow \frac{1}{16}\left[\mathrm{\Sigma }(QA{)}^2-\mathrm{\Sigma }(PA{)}^2\right]=\frac{1}{16}\left(16\times {16}^2-12\times {16}^2\right)=\frac{1}{16}\left(4\times {16}^2\right)=64$.