We claim that the fixed point is $O$, the circumcenter of $\triangle ABC.$ To show this, we'd just have to show that $(\triangle AMN), (\triangle AXY), (\triangle ABC)$ are coaxial. Since they share the point $A$, we just need to show that they share another point. Invert at $A$ with arbitrary radius, and let $P'$ denote the image of point $P.$ We require that $B'C', M'N', X'Y'$ are concurrent. Let $\gamma_b', \gamma_c'$ denote the images of $\gamma_b, \gamma_c$ under this inversion. Since $M, N$ lie on segments $AB, AC$, $\gamma_b'$ and $\gamma_c'$ are also circles. It's easy to see that $X', Y'$ are the centers of $\gamma_b', \gamma_c'$ respectively. We claim that $X'Y', M'N', B'C'$ concur at the exsimilicenter of $\gamma_b'$ and $\gamma_c'$, say $Z.$ It's clear that $Z \in X'Y'.$ Since $X'M' || O_b B$ (as $MXO_bB$ is cyclic) and $Y'N' || O_c C$ (as $NY O_c C$ is cyclic), we see that $X'M' || Y'N'.$ Hence, the homothety centered at $Z$ sending $\gamma_b'$ to $\gamma_c'$ also sends $M'$ to $N'$, implying that $Z \in M'N'.$ So let's just check that $Z \in B'C'.$ Notice that $(\triangle AB'C')$ is the image of line $BC$ under the inversion, and therefore is tangent to both $\gamma_b'$ and $\gamma_c'$ (inversion preserves tangency). By Monge on $\gamma_b', \gamma_c', (\triangle AB'C')$, we then get that $B', C', Z$ are collinear, as desired. Hence, we've shown that $Z = B'C' \cap X'Y' \cap M'N'$, so inverting back implies that $Z'$ is a common point of $(\triangle AMN), (\triangle AXY), (\triangle ABC).$ Therefore, the circles are indeed coaxial and so the line through the centers of $(\triangle AMN), (\triangle AXY)$ always passes through $O.$ $\square$