Let $H$ be the intersection of $B_1C_2$ and $B_2C_1$. We will show that $H$ is the center of the homothety. Apply Pascal's theorem give us $H, P, Q$ collinear. Consider this configuration in the complex plane, and let $(O)$ be the unit circle. Let $F$ and $G$ be the center of $\Gamma_2$ and $\Gamma_1$, then $$h=\frac{b_{1}c_{2}(b_{2}+c_{1})-b_{2}c_{1}(b_{1}+c_{2})}{b_{1}c_{2}-b_{2}c_{1}}$$$$g=\frac{2b_{1}c_{1}}{b_{1}+c_{1}}, f=\frac{2b_{2}c_{2}}{b_{2}+c_{2}}$$Since $$\frac{h-g}{h-f}=\frac{(b_{1}-c_{1})(b_{2}+c_{2})}{(b_{2}-c_{2})(b_{1}+c_{1})} \in \mathbb{R},$$we have $H,F,G$ collinear Also, we have $$g-c_{1}=\frac{c_{1}(b_{1}-c_{1})}{b_{1}+c_{1}}, f-c_{2}=\frac{c_{2}(b_{2}-c_{2})}{b_{2}+c_{2}}$$Therefore $$\left|\frac{h-g}{h-f}\right|=\left|\frac{g-c_{1}}{f-c_{2}}\right|$$or $$\frac{HG}{HF}=\frac{GC_{1}}{FC_{2}}$$From here we conclude that there exist a homothety with center $H \in PQ$ sends $\Gamma_1$ to $\Gamma_2$.

The complex solution above is amazing! Here's a purely geometric solution (haven't used pole-polar in 10 years lol), which is normal, even though it took me so much time to find... Let $R = B_1C_2 \cap B_2C_1$ which helpfully lies on $PQ$ by Pascal. In fact, we can now remove $A_1$, $A_2$, $P$, $Q$ entirely from the diagram. Let $O_1$ and $O_2$ be the centers of the two circles. Note that the polar of $O_1$ is $B_1C_1$, the polar of $O_2$ is $B_2C_2$ and the polar of $R$ passes through $B_1C_1 \cap B_2C_2$ by Brocard's theorem, so the three polars are concurrent - hence $O_1, O_2, R$ are collinear. (Ok, this also follows by Pascal on the degenerates $C_2B_1B_1B_2C_1C_1$ and $C_1B_2B_2B_1C_2C_2$. Alternatively, invert with respect to the main circle and notice that the images lie on the circle with diameter $O(B_1C_1\cap B_2C_2)$ by Brocard's theorem.) Now to justify the homothety, it suffices to show $\frac{RO_1}{RO_2} = \frac{O_1C_1}{O_2C_2}$, since the latter are the radii of the circles. Equivalently, $\frac{O_1B_1}{O_1R} = \frac{O_2C_2}{O_2R}$, i.e. $\frac{\sin \angle B_1RO_1}{\sin \angle O_1B_1R} = \frac{\sin \angle O_2RC_2}{\sin \angle RC_2O_2}$ by the Sine Law. The latter follows by $\angle O_1B_1R = \angle B_2B_1R - \angle B_2B_1O_1 = \angle B_2A_2C_2 - \angle B_2A_2B_1 = \angle B_1C_2O_2$ and the analogous for the others.

Well, this is really similar to Bulgaria RMM TST 2024 G3 and the RMM TST was held a few days ago, so that's pretty interesting. We have to show that the insimilicenter of $\Gamma_1, \Gamma_2$ lies on $PQ$. Clearly $B_1C_2 \cap B_2C_1=X \in PQ$ by Pascal on $C_2B_1A_1C_1B_2A_2$, so we claim this is the insimilicenter. Let $\Omega_1$ be the circle tangent to $OB_1, OC_2$ at $B_1, C_2$ (exists since $OB_1=OC_2$) and let $\Omega_2$ be the circle tangent to $OB_2, OC_1$ at $B_2, C_1$. By Monge on $\Omega_1, \Gamma_1, \Gamma_2$, the insimilicenter lies on $B_1C_2$ and by Monge on $\Omega_2, \Gamma_1, \Gamma_2$, the insimilicenter lies on $C_1B_2$. Thus, $X$ is the insimilicenter. Remarks: The idea of applying twice Monge comes from the official solution of the above-mentioned problem, so credits to @Strudan_Borisov, who is the proposer of this problem. I think my approach from the problem above with inversion at $X$ might also work as well, but it's probably longer.

One may recognise the Brocard configuration for cyclic quadrilaterals (I prefer to call it so) and observe that the problem is equivalent to the following: Given a convex quadrilateral $ABCD$, which is inscribed in a circle centred at $O$. Let $AC\cap BD=R(.)$ and consider its polar line w.r.t. $(O)$ that contains points $AB\cap CD=Q(.)$, $AD\cap BC=P(.)$. If we define points $X$ and $Y$ as poles of lines $AC$ and $BD$ w.r.t. $(O)$, respectively, then clearly both lie on $PQ$. We are asked to show that \[ \frac{PX}{PY} = \frac{XC}{YB} \quad \text{or/and} \quad \frac{QX}{QY} = \frac{XC}{YB}, \]in other words, the points $P$ and $Q$ are exsimilicenter and insimilicenter of circles centered at $X$ and $Y$ with radii $XC$ and $YB$, respectively. The bundle $\{P, X, Q,Y\}$ is harmonic, so it might be appropriate to ask to show that $\frac{PX\cdot QX}{\mathrm{Pow} (X)} = \frac{PY\cdot QY}{\mathrm{Pow} (Y)}$ in the given figure. Alternatively, showing $\frac{PX}{PY} = \frac{XC}{YB}$ can be done simply by considering triangles $PXC$ and $PYB$ along with the Sine Law .

Attachments:

We use the following well-known result: If there exists an inversion from some point that swaps two circles, then this point is a center of a similitude of those two circles. Let $R = B_1C_2 \cap C_1B_2$. Pascal's theorem directly implies that $R \in PQ$. Consider the inversion $\Phi$ from $R$ with radius $\sqrt{RB_2 \cdot RC_1}$. We will show that $\Phi$ swaps $\Gamma_1$ and $\Gamma_2$. Cleary, $\Phi$ swaps $B_1 \leftrightarrow C_2$ and $B_2 \leftrightarrow C_1$. We need another pair of points getting swapped and we'll be done. Let $X = B_1C_2 \cap \Gamma_1, Y = B_1C_2 \cap \Gamma_2$. If we show that $C_1B_2XY$ is cyclic, then power of a point $R$ will imply that $X \leftrightarrow Y$. However, this is just an angle-chase. $$\angle{B_2C_1X} = \angle{OC_1X} - \angle{OC_1B_2} = \pi - \angle{C_1B_1X} - \angle{OB_2C_1} = \angle{C_1B_1C_2} - \angle{OB_2C_1} = \angle{C_1B_2C_2} - \angle{OB_2C_1} = \angle{OB_2C_2} = \angle{OC_2B_2} = \angle{B_2YX}$$and we are done. $\ \square$

Attachments:

My solution from the contest Let $O_1$ and $O_2$ be the circumcenters of $\Gamma_1$ and $\Gamma_2$ and let $r_1$ and $r_2$ be their radii. Let $Z = C_2B_1 \cap B_2C_1$, by Pascal on $A_2C_2B_1A_1C_1B_2$ we get that $P,Q$ and $Z$ are collinear. It is enough to show that $Z$ is the insimilicenter of $\Gamma_1$ and $\Gamma_2$. Since $O_1,O_2$ and $Z$ all lie on the polar of $C_1B_1 \cap B_2C_2$ with respect to $\Omega$, we have that they are collinear. Now let $\alpha,\beta,\gamma,\delta$ be the angles $\angle O_1C_1Z,\angle ZC_2O_2,\angle O_1ZC_1$ and $\angle C_2ZO_2$, respectively and let $x = O_1Z$ and $y = O_2Z$. Note that $\angle ZB_1O_1 = \beta$ and $\angle ZB_2O_2 = \alpha$. From the law of sines on $\triangle O_1ZC_1$ and $\triangle O_1ZB$ we get $$\frac{x}{\sin{\alpha}} = \frac{r_1}{\sin{\gamma}} \text{ and } \frac{x}{\sin{\beta}} = \frac{r_1}{\sin{\delta}}$$respectively. Multiplying the two, we have $$\frac{x^2}{\sin{\alpha}\cdot\sin{\beta}} = \frac{r_1^2}{\sin{\gamma}\cdot \sin{\delta}}$$Analogously for $\triangle ZC_2O_2$ and $ZB_2O_2$ $$\frac{y^2}{\sin{\alpha}\cdot\sin{\beta}} = \frac{r_2^2}{\sin{\gamma}\cdot \sin{\delta}}$$Finally, dividing the two results we obtain $\frac{x^2}{y^2} = \frac{r_1^2}{r_2^2} \Rightarrow \frac{x}{y} = \frac{r_1}{r_2}$ so $Z$ is indeed the insimilicenter of $\Gamma_1$ and $\Gamma_2$.