Since the angles $\angle OXP$ and $\angle OYQ$ are both right, the circles $OXP$ and $OYQ$ cross again at a point $R$ on $PQ$, and $OR$ is perpendicular to $PQ$. Letting $A_1C_1$ and $A_2B_2$ cross at $S$, the conclusion then follows at once from the two facts below:
(1) $S$ lies on $OR$; and
(2) $OS$ is perpendicular to $A_1A_2$.

To prove (1), invert from $O$ with power $-OB_1 \cdot OB_2$. Under this inversion, $\gamma_1$ and $\gamma_2$ correspond to one another, and so do the points in each of the pairs $(A_1, Y)$, $(B_1, B_2)$, $(C_1, C_2)$ and $(X, A_2)$.

Clearly, $B_1$ lies on the circle $OXP$, so this latter is mapped to $A_2B_2$. Similarly, the circle $OYQ$ (through $C_2$) is mapped to $A_1C_1$.
Consequently, $R$ is mapped to $S$. This establishes (1).

To prove (2), let $A_1A_2$ cross $B_1B_2$ and $C_1C_2$ at $D_1$ and $D_2$, respectively; let $O_1O_2$ cross $A_1C_1$ and $A_2B_2$ at $X_1$ and $X_2$, respectively; and let $A_1C_1$ cross $O_1D_1$ at $Y_1$, and $A_2B_2$ cross $O_2D_2$ at $Y_2$. The argument hinges on the following two facts:
(3) $X_1$ and $Y_1$ both lie on the circle $\omega_1$ on diameter $OD_1$; similarly, $X_2$ and $Y_2$ both lie on the circle $\omega_2$ on diameter $OD_2$; and
(4) $X_1, X_2, Y_1, Y_2$ all lie on a circle $\omega$.
Assume (3) and (4) to establish (2) as follows: $X_1Y_1$, i.e., $A_1C_1$, is the radical axis of $\omega$ and $\omega_1$, and $X_2Y_2$, i.e., $A_2B_2$, is the radical axis of $\omega$ and $\omega_2$. Hence $S$ is the radical centre of the three circles. As such, $S$ lies on the radical axis of $\omega_1$ and $\omega_2$. These two circles cross again at the orthogonal projection $O'$ of $O$ on $D_1D_2$, i.e., on $A_1A_2$, and $OO'$ is their radical axis. Consequently, $S$ lies on $OO'$ and (2) follows.

We now turn to prove (3) and (4).
To prove (3) only $X_1$ and $Y_1$ are dealt with. It is sufficient to show that the angles $\angle O_1X_1D_1$ and $\angle OY_1O_1$ are both right.
To prove that the angle $\angle O_1X_1D_1$ is right, we show that $X_1$ lies on the circle on diameter $O_1D_1$. Clearly, $A_1$ and $B_1$ both lie on this circle, so it is sufficient to show that $\angle O_1A_1X_1 = \angle O_1B_1X_1$. Now, $\angle O_1A_1X_1 = \angle O_1A_1C_1 = \angle O_1C_1A_1 = \angle O_1C_1X_1$, since $O_1A_1 = O_1C_1$ (they both are radii of $\gamma_1$). On the other hand, $C_1$ and $B_1$ are reflexions of one another in $OO_1$, so $\angle O_1C_1X_1 = \angle O_1B_1X_1$. Consequently, $\angle O_1A_1X_1 = \angle O_1B_1X_1$, as desired.
The proof that the angle $\angle OY_1O_1$ is right is quite similar: This time we show that $Y_1$ lies on the circle on diameter $OO_1$. Clearly, $B_1$ and $C_1$ both lie on this circle, so it is sufficient to show that $\angle O_1B_1Y_1 = \angle O_1C_1Y_1$. Notice that $B_1$ and $A_1$ are reflexions of one another in $O_1D_1$, to write $\angle O_1B_1Y_1 = \angle O_1A_1Y_1$. Refer now again to $O_1A_1 = O_1C_1$, to write $\angle O_1A_1Y_1 = \angle O_1A_1C_1 = \angle O_1C_1A_1 = \angle O_1C_1Y_1$ and conclude that $O_1B_1Y_1 = \angle O_1C_1Y_1$. This establishes (3).

Finally, proving (4) requires more work. We will show that $\angle X_1Y_1Y_2+\angle X_1X_2Y_2 = 180^\circ$. Clearly, $\angle X_1Y_1Y_2 = 180^\circ-(\angle O_1Y_1X_1+\angle D_1Y_1Y_2)$, so it is sufficient to prove that
$$\angle X_1X_2Y_2 = \angle O_1Y_1X_1 + \angle D_1Y_1Y_2. \quad (*)$$
Deal with each angle in the right-hand member separately. By (3), $\angle O_1Y_1X_1 = \angle O_1OD_1$, and an obvious angle chase yields $\angle O_1OD_1 = \angle O_1OB_1 = \angle O_2OB_2 = \angle O_2OC_2 = \angle O_2OD_2$. Hence $\angle O_1Y_1X_1 = \angle O_2OD_2$.
To deal with $\angle D_1Y_1Y_2$, let $O_1D_1$ and $O_2D_2$ cross at $Z$. By (3), the angles $\angle OY_1D_1$ and $\angle OY_2D_2$ are both right, so $OY_1ZY_2$ is cyclic, and hence $\angle D_1Y_1Y_2 = \angle ZY_1Y_2 = \angle ZOY_2$.
Thus, (*) now reads
$$\angle X_1X_2Y_2 = \angle O_2OD_2 + \angle ZOY_2. \quad (**)$$
Our purpose now is to replace $\angle ZOY_2$ by a more tractable angle. To this end, notice that $O_1D_1$ is the internal angle bisectrix of $\angle A_1D_1B_1$, and $O_2D_1$ is the internal angle bisectrix of $\angle A_2D_1B_2$, so $O_1D_1$ and $O_2D_1$ are perpendicular; that is, $O_2D_1$ is the $O_2$-altitude of the triangle $ZO_1O_2$. Similarly, $O_1D_2$ is the $O_1$-altitude of the triangle $ZO_1O_2$, so it crosses $O_2D_1$ at the orthocentre $H$ of this triangle.
On the other hand, $O_2D_1$ and $O_1D_2$ are clearly the internal angle bisectrices of the triangle $OD_1D_2$ at $D_1$ and $D_2$, respectively. Hence $OH$ is the internal angle bisectrix of $\angle D_1OD_2$. As such, it is perpendicular to external bisectrix $O_1O_2$ of this angle. Recalling that $H$ is the orthocentre of the triangle $ZO_1O_2$, it follows that $OH$ is the $Z$-altitude of this triangle.
Hence $\angle ZOY_2 = \angle HOY_2$, and (**) now reads
$$\angle X_1X_2Y_2 = \angle O_2OD_2 + \angle HOY_2.$$
To prove this, split $\angle X_1X_2Y_2 = \angle X_1X_2D_2 + \angle D_2X_2Y_2$. The angle $\angle X_1X_2D_2 = \angle OX_2D_2$ is right, by (3). As such, $\angle X_1X_2D_2 = \angle HOX_2$, since $OH$ is perpendicular to $O_1O_2$. Split further $\angle HOX_2 = \angle HOY_2 + \angle Y_2OX_2$, to write $\angle X_1X_2D_2 = \angle HOY_2 + \angle Y_2OX_2$. By (3), $\angle D_2X_2Y_2 = \angle D_2OY_2$, so $\angle X_1X_2Y_2 = \angle HOY_2 + \angle Y_2OX_2 + \angle D_2OY_2 = \angle HOY_2 + \angle D_2OX_2$, as desired. This establishes (4) and completes the proof.