Let $D_1$ and $D_2$ be points on the circles such that $C_1D_1$ and $C_2D_2$ are diameters. We want to show that the reflection in the line $C_1C_2$ of the circumcircle of $\triangle C_1BC_2$ is the circumcircle of $\triangle C_1AC_2$. Equivalently we can show that the reflection of $B$ in the line $C_1C_2$ is on the circumcircle of $\triangle C_1AC_2$, which is equivalent to
$$\angle C_1BC_2 = 180^\circ - \angle C_1AC_2.$$
Instead of using reflection in the line $C_1C_2$, we could also argue with the extended Sine-Rule, which tells us that the circumradius of $\triangle C_1AC_2$ is equal to $|C_1C_2|/2 \sin(\angle C_1AC_2)$ and the circumradius of $\triangle C_1BC_2$ is equal to $|C_1C_2|/2 \sin(\angle C_1BC_2)$. We then see that these circles have the same radius if $\sin(\angle C_1AC_2) = \sin(\angle C_1BC_2)$, which follows when we have shown $\angle C_1BC_2 = 180^\circ - \angle C_1AC_2$.

$$\angle C_1AC_2 = \angle C_1D_1B + \angle BD_2C_2.$$
As $C_1D_1$ and $C_2D_2$ are parallel, $\angle C_1D_1D_2 + \angle D_1D_2C_2 = 180^\circ$. Using the previously obtained equation this yields
$$\begin{aligned} \angle C_1AC_2 &= \angle C_1D_1B + \angle BD_2C_2 = 180^\circ - (\angle BD_1D_2 + \angle D_1D_2B) \\ &= \angle D_1BD_2. \end{aligned}$$
Since $C_1D_1$ and $C_2D_2$ are diameters, $\angle C_1BD_1 = 90^\circ = \angle C_2BD_2$ and we finally obtain $\angle C_1BC_2 = 180^\circ - \angle D_1BD_2 = 180^\circ - \angle C_1AC_2$, as desired.