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

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 $F \in A M)$.

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