The points $P,Q,O$ are collinear(on the perpendicular bisector of $AB$) $S_{AOBQ}=\frac{\mid AB\mid\cdotp\mid OQ\mid}{2},S_{AOBP}=\frac{\mid AB\mid\cdotp\mid OP\mid}{2}\Rightarrow\mid OP\mid\cdotp\mid OQ\mid=\frac{S_{AOBQ}\cdotp S_{AOBP}}{R^{2}\cdotp sin^{2}C}$ But,$m\left(\angle PBA\right)=m\left(\angle PAB\right)=90\textdegree-m\left(\angle A\right)\Rightarrow m\left(\angle APB\right)=2m\left(\angle A\right)$ If $R_{1}$ is the circumradius of circle with the center in P,in the $\triangle APB$,we have: $4R^{2}sin^{2}C=2R_{1}^{2}-2R_{1}^{2}cos2A\Rightarrow R_{1}=\frac{RsinC}{sinA}$ Then:$S_{APBO}=S_{APB}+S_{AOB}=\frac{R^{2}sin2C}{2}+\frac{R^{2}sin^{2}Csin2A}{sin^{2}A}=\frac{R^{2}sinC\cdotp sinB}{sinA}$ Similarly.$S_{AQBO}=\frac{R^{2}sinC\cdotp sinA}{sinB}\Rightarrow S_{APBO}\cdotp S_{AQBO}=R^{4}sin^{2}C\Rightarrow\mid OP\mid\cdotp\mid OQ\mid=R^{2}$.