Circumscribed circles of the triangles I'm interested in the following problem

Let $\omega$ be a circle with O the center of the circle and I a straight line. Perpendicular from O on the line I cut $\omega$ into A and B. Let P and Q be two points on the $\omega$ and let $PA\cap I=X_1$,$PB\cap I=X_2$, $QA\cap I=Y_1$, $QB\cap I=Y_2$. Prove that the circumscribed circles of the triangles $AX_1 Y_1$ and $AX_2 Y_2$ intersect a second time at a point on $\omega$.

My attempt. I saw that the points $P$ and $Q$ are mobile so I tried finding a projective function and then applying the moving points method but I am not very good at this method. I also tried to use an inversion but I don't think that it would work.
Any ideas please!
 A: Lemma. Let $\ell$ be a line and $\mathcal C$ be a circle with center $\mathcal C_O$. Let the line passing through $\mathcal C_O$ perpendicular to $\ell$ intersect $\mathcal C$ at $\{\mathcal C_1, \mathcal C_2\}$. Let $(\ell_1,\ell_2)\in\ell^2$ be two points on $\ell$ such that $\ell_i\mathcal C_1\cap \mathcal C=\mathcal P_i(\neq \mathcal C_1)$ for $i\in\{1,2\}$. Then $\{\mathcal P_1,\mathcal P_2,\ell_1,\ell_2\}$ are concyclic.
Proof. Note that
$$\angle \ell_2\ell_1\mathcal C_1=90-\angle \mathcal P_1\mathcal C_1\mathcal C_2=\angle \mathcal C_1\mathcal C_2\mathcal P_1=\angle \mathcal C_1\mathcal P_2\mathcal P_1\implies \{\mathcal P_1,\mathcal P_2,\ell_1,\ell_2\}\text{ are concyclic.}$$
$$\tag*{$\square$}$$
Let $\{Y_1B,X_1B\}\cap \omega:=\{E,D\}$. Note that $X_1P\perp BX_2$ and $BA\perp \ell\implies A$ is orthocenter of $\triangle X_2BX_1$ and as $AD\perp BX_1$, we get $\{X_2-A-D\}$ are collinear. Similarly, $\{Y_2-A-E\}$ are collinear. Thus, by our lemma, $X_1Y_1ED$ and $Y_2X_2ED$ are cyclic.
We claim that $\{DE,X_2Y_2,PQ\}$ concur at a point $C$. This is obvious by pascal's theorem on $BPQADE$.
Now, note that by power of point, we get
$$CQ\cdot CP=CD\cdot CE=CY_2\cdot CX_2=CX_1\cdot CY_1$$
Thus, point $C$ has equal powers with respect to $\{\omega, \odot(AX_1Y_1),\odot(AX_2Y_2)\}$ and as $C\not\equiv A$, these three circles must be coaxial completing the proof.
$$\tag*{$\blacksquare$}$$

