Prove that the circumradius of $\triangle PAB$ is constant

Question

Given two circle $\omega_1$ and $\omega_2$ which intersect at points $X$ and $Y$. Let $P$ be an arbitrary point on $\omega_1$. Suppose that the lines $PX$ and $PY$ meet $\omega_2$ again at $A$ and $B$ respectively. Prove that the circumcircles of all triangles $PAB$ have equal radius.

What I tried is not worth adding here.

Answer 1

Setup #1 Initially, we have $\triangle PAB$ with P on $\omega_1$ and chord AB of $\omega_2$.

Setup #2 When P moves along $\omega_1$ to P’, we have the corresponing image chord A’B’ traced out.

The keypoints are:-

(1) $\angle P$ is invariant (i. e. $\angle P = \angle P’$); and

(2) The length of the chord AB is invariant (i. e. AB = A’B’).

In addition, if R is the circum-radius of $\triangle PAB$, then $AB = 2R \sin \angle P$.

The proof of all the above can be found in my explanation to the question in here.

Suppose that the circum-radius of $\triangle P’A’B’$ is $R’$. Then, $A’B’ = 2R’ \sin \angle P’$.

Combining all of the above, we have R = R’.

Answer 2

Hint: The circumradius $R$ of triangle $PXY$ satisfies $2R = |XY|/\sin \angle XPY$. Because $\angle XPY = \angle APB$ is constant, it is sufficient to show that $|XY|$ is constant as well. In fact $|XY|$ being constant again comes from $\angle APB$ being constant.