A geometry problem by Ricky Huang

$\overline {OB}$ is the median of $\triangle {OAP}$

So, $|\overline {OA}|^2+|\overline {OP}|^2=2|\overline {BP}|^2+2|\overline {OB}|^2$

Let the radius of the circle be $r$

$\triangle {OCP}$ is right angled isosceles. So, $|\overline {OP}|=r\sqrt 2$

Hence, $r^2+2r^2=2(\sqrt 2) ^2+2r^2$

$\implies r=(\sqrt 2) ^2=\boxed 2$ .

It is interesting to note that $|\overline {OP}|=|\overline {AP}|=2\sqrt 2$

Using secant-tangent theorem, we have: $CP^2 =BP\cdot AP=\sqrt 2(2 \sqrt 2) = 4 \implies CP =2$ .

Since $\angle CPD =45^\circ$ , $OP=2\sqrt 2$ . Let the radius of the circle be $r$ . Then we have:

$\begin{aligned} EP \cdot DP & = CP^2 \\ (2\sqrt 2-r)(2\sqrt 2 +r) & = 2^2 \\ 8-r^2 & =4 \\ r^2 & =4 \\ \implies r & = \boxed 2 \end{aligned}$