Solution suggested by the Student Alzubair Habibullah. Because $M$ is the midpoint of $O_{1} O_{2}$, we have from Apollonius' theorem
$$4 M S^{2} = 2 S O_{1}^{2} + 2 S O_{2}^{2} - O_{1} O_{2}^{2}$$
Because $S$ is the midpoint of segment $A X$, segment $O_{1} S$ is perpendicular to segment $A X$ and therefore $S O_{1}^{2} = O_{1} A^{2} - A S^{2}$.
On the other hand, we have from the power of point $S$ with respect to circle $\omega_{2}$, $S O_{2}^{2} = S A \cdot S Z + O_{2} A^{2} = S A^{2} + S A \cdot A Z + O_{2} A^{2}$.
We deduce that
$$4 M S^{2} = 2 O_{1} A^{2} + 2 O_{2} A^{2} + 2 S A \cdot A Z - O_{1} O_{2}^{2} = 2 O_{1} A^{2} + 2 O_{2} A^{2} + X A \cdot A Z - O_{1} O_{2}^{2}$$

We deduce in a similar way that
$$4 M T^{2} = 2 O_{2} A^{2} + 2 O_{1} A^{2} + W A \cdot A Y - O_{2} O_{1}^{2}$$
Therefore
$$4\left(M S^{2} - M T^{2}\right) = X A \cdot A Z - W A \cdot A Y$$
This proves that $M S = M T$ if and only if $X A \cdot A Z = W A \cdot A Y$, which is equivalent to saying that $X, Y, Z$ and $W$ are concyclic.