Let $O=(0,0)$, $A= (0,a)$ and $B= (b,0)$, where $0<a<b$ are reals. Let $\Gamma$ be a circle with diameter $\overline{AB}$ and let $P$ be any other point on $\Gamma$. Line $PA$ meets the $x$-axis again at $Q$. Prove that $\angle BQP = \angle BOP$.
Here is my interpretation of the diagram
Is it incorrect to assume that $\Gamma$ is the circumcircle of $\triangle AOB$? Can anyone tell me what's the error?