Given are two circles $\omega_1,\omega_2$ which intersect at points $X,Y$. Let $P$ be an arbitrary point on $\omega_1$. Suppose that the lines $PX,PY$ meet $\omega_2$ again at points $A,B$ respectively. Prove that the circumcircles of all triangles $PAB$ have the same radius.
I can observe that the $\angle P$ and the chord $XY$ remains constant throughout the problem. What are the other key observations that can be noted from the given problem?
My diagram looks like this:
Let the centres of circles $\omega_1$ and $\omega_2$ be $O_1$ and $O_2$ respectively.
Let $\angle XO_1Y$ be $2\alpha$ and $\angle XO_2Y$ be $2\beta$, both of which are constant no matter where $P$ is.
As you have correctly identified, since the angle at the centre is twice the angle at the circumference, $\angle XPY=\alpha$ which is constant.
For the same reason we get that $\angle XBY=\beta$ which is also constant.
Therefore $\angle PXB=180^\circ-\alpha-\beta$ and so $\angle AXB=180^\circ-\angle PXB=\alpha+\beta$.
Once again due to angle at the centre is twice the angle at the circumference, we get that $\angle AO_2B=\frac12(\alpha+\beta)$.
$O_2A=O_2B$ is the radius of $\omega_2$ and so is constant, and $\angle AO_2B=\frac12(\alpha+\beta)$ is also constant, so all possible triangles $AO_2B$ are congruent, which means that length $AB$ is constant.
Using the full sine rule, we get that $$2R=\frac{AB}{\sin\alpha}$$ where $R$ is the circumradius of triangle $PAB$, but since length $AB$ and $\alpha$ are constant no matter where $P$ is, we get that the circumcircles of all triangles $PAB$ have the same radius.