Proof: Let $O_1$ and $O_2$ be the centers of circles $\omega_1$ and $\omega_2$, respectively. By symmetry, we know that $\angle S_1O_1Y_1 = \angle S_1O_1X_1$. Since $M$ is the midpoint of $AB$, we have $O_1M \perp S_1M$. Due to the tangent lines, we have $O_1X_1 \perp S_1X_1$ and $O_1Y_1 \perp S_1Y_1$. Therefore, points $S_1$, $O_1$, $X_1$, $M$, and $Y_1$ are concyclic. Similarly, points $S_2$, $O_2$, $X_2$, $M$, and $Y_2$ are concyclic. From this, we can conclude that $\angle S_1MY_1 = \angle S_1O_1Y_1 = \angle S_1O_1X_1 = \angle S_1MX_1 = \angle S_2MX_2 = \angle S_2O_2X_2 = \angle S_2O_2Y_2 = \angle S_2MY_2$. Therefore, $M$, $Y_1$, and $Y_2$ are collinear.