Given $2$ circles $W_1$ and $W_2$ which intersect at points $X$,$Y$. Let $P$ be an arbitrary point on $W_1$. Suppose the lines $PX$,$PY$ meet $W_2$ again at points $A$,$B$ respectively. Prove that the circumcircles of all triangles $PAB$ have the same radius.
MY TRY:- I tried to use coordinate geometry, but it didn't work. Please help me.
Let $O_1$ and $O_2$ be the centers of circles $W_1$ and $W_2$ respectively. Denote by $$\alpha_1 = \angle \, XPY = \angle \, APB = \angle \, XPB = \frac{1}{2} \, \text{arc}_{W_1}(XY)$$ and $$\alpha_2 = \angle \, XBY = \angle \, XAY = \angle \, XBP = \frac{1}{2} \, \text{arc}_{W_2}(XY)$$ Notice that these two arcs, and therefore the angles $\alpha_1$ and $\alpha_2$, are independent of the choice of the point $P$ and the consequent location of the chord $AB$.
Then $$\text{arc}_{W_2}(AB) =2 \, \angle \, AXB = 2 \, (\angle \, XPB + \angle \, XBP ) = 2\,(\alpha_1 + \alpha_2)$$ Hence, the arc $\text{arc}_{W_2}(AB)$ on circle $W_2$ which does not contain points $X, Y$ is of constant length $2\,(\alpha_1 + \alpha_2)$ for any choice of the point $P$ on $W_1$. Therefore the chord $AB$ is also of constant length for any choice of the point $P$ on $W_1$. The circle circumscribed around any configuration $PAB$ has radius $$R = \frac{AB}{2 \, \sin(\angle \, APB)} = \frac{AB}{2 \, \sin(\alpha_1)}$$ so it is independent on the choice of the point $P$ on $W_1$.