Let $\Omega_1\cap AX,AY=D,D_1$ and $\Omega_2\cap BX,BY=C,C_1$ and $T=AB\cap CD$. $O_i$ be center of $\Omega_i$. As $X$ and $Y$ lie on radical axis of $\Omega_1$ and $\Omega_2$, $ABCD$ and $ABC_1D_1$ are cyclic $\implies \angle ADD_1=\angle ABX$ and $\angle AD_1D=\angle ABC_1\implies \angle DAD_1=\angle CBC_1\implies \angle DO_1D_1=\angle C_1O_2C$. Also we have $TD\cdot TC=TA\cdot TB=TD_1\cdot TC_1\implies \frac{TD}{TC_1}=\frac{TD_1}{TC}$, which means $T$ is the center of homothety that maps $DD_1$ to $C_1C$. As $\triangle O_1DD_1\sim \triangle O_2C_1C$ and as $O_1,O_2$ lie on the same side of $DC$, this homothety maps $O_1$ to $O_2$. So $T-O_1-O_2$ are collinear, which implies $T$ lies on perpendicular bisector of $PQ$.