Circles and tangent. 
The circle $A$ touches the circle $B$ internaly at $P$. The centre $O$ of $B$ is outside $A$. Let $XY$ be a diameter of $B$ which is also tangent to $A$. Assume $PY>PX$. Let $PY$ intersect $A$ at $Z$. If $YZ=2PZ$, what is the magnitude of angle $PYX$ in degrees?

I know that angle $XPY$ will be $90^\circ$. But I dont know how to use the information $YZ=2PZ$. I tried using the property that the angle subtended by chord on circumference is half of angle subtended by the chord on the center, but reached nowhere.
Any help would be appreciated.
 A: 
Looking at the diagram above, note that $\triangle ZPR$ and $\triangle YPO$ are similar since they are isosceles (each with two radii for shorter sides) and share a common angle. From this and $YZ=2ZP$ we have $OR=2RP=2RS$, so $\angle ROS=30^\circ$ ($\triangle ORS$ being half an equilateral triangle). $\angle ROS=\angle POX=2\angle PYX$ by the central/inscribed angle theorem, so $\angle PYX=15^\circ$.
Nothing beyond basic geometry was needed here.
A: Let $PX$ intersect circle $A$ at $W$. Then $\triangle PWX$ and $\triangle PXY$ are homothetic with center $P$(that is, they are similar). If we set $PZ=1$, we have $YZ=2$, and if $XY$ is tangent to $A$ at $T$ then $TY=\sqrt{6}$ by power of a point. Similarly by homothety, if we set $PW=x$, then $XW=2x, XT=\sqrt{6}x$. Since $\triangle PXY$ is a right triangle, we have $PX^2+PY^2=XY^2$, that is, 
$(3(x+1))^2+3^2=(\sqrt{6}(x+1))^2$. Solving this leads to $x=2-\sqrt{3}$ ($x=2+\sqrt{3}$ cannot be a solution here because $O$ is outside of $A$). Since $\tan15^{\circ}=2-\sqrt3$, we have $\angle PYX=15^{\circ}$.
