Let the circles be $(O_1,R_1)$ and $(O_2,R_2).$ Let $\theta=\angle PO_1O_2$ and $\phi=\angle PO_2O_1,$ with $\theta$ and $\phi$ given by construction. Let $A$ be on circle $(O_1,R_1)$ and $B$ on $(O_2,R_2).$ Let $x=\angle O_1PA$ and $y=\angle O_2PB.$ It is obvious that $x+y=\theta+\phi<180^\circ,$ $AP=2R_1\cos x,$ and $PB=2R_2\cos y=2R_2\cos (\theta+\phi-x).$ Then $AP\cdot PB=4R_1R_2\cos x\cdot \cos(\theta+\phi-x)=2R_1R_2\left[\cos(\theta+\phi)+\cos(\theta+\phi-2x)\right].$ Therefore $AP\cdot PB=\text{max}$ when $2x=\theta+\phi$ and $x=y=(\theta+\phi)/2.$ Thus, $APB$ is the external angle bisector of $\triangle O_1PO_2$ at $P,$ which can be constructed easily.