Shortlist JBMO, 2008

Solution:
Consider the line $\rho$ tangent to $\gamma$ at $A$, and take the points $K = A M \cap X Y$, $L = \rho \cap X M$, and $F = O A \cap X Y$.

(Remark: Moving $M$ into its reflection with respect to the line $O A$ will move $X Y$ into its reflection with respect to $O A$. These old and the new $X Y$ meet on $O A$, hence it should be clear that the fixed point must be $F$.)

Since $\angle L M A = \angle F Y A$ and $\angle Y A F = \angle L A M = 90^\circ$, it follows that triangles $F A Y$ and $L A M$ are similar, therefore $\angle A F Y = \angle A L M$, hence the quadrilateral $A L X F$ is cyclic. But then $\angle A F L = \angle A X L = 90^\circ$, so $L F \perp A F$, hence $L F \parallel \delta$.

Now, $\rho$ is the radical axis of circles $\gamma$ and $A$ (consider $A$ as a circle of center $A$ and radius $0$), while $X M$ is the radical axis of circles $\gamma$ and $\Gamma$, so $L$ is the radical center of the three circles, which means that $L$ lies on the radical axis of circles $\Gamma$ and $A$. From $L F \perp O A$, where $O A$ is the line of the centers of the circles $A$ and $\Gamma$, and $F \in X Y$, it follows that $F$ is (the) fixed point of $X Y$.

(The degenerate two cases when $M \in O A$, where $X \equiv M$ and $Y \equiv A$, also trivially satisfy the conclusion, as then $F \in A M$.)