$WLOG$, Let $\Gamma_2$ be not smaller than $\Gamma_1$ First, we define some points Let $X,Y$ be the intersection of the external angle bisector of $\angle C_1BC_2$ with $\Gamma_1,\Gamma_2$ distinct from $B$ $Z=XC_1 \cap YC_2$ $K$ be the intersection of the perpendicular to $XZ$ through $X$ and the perpendicular to $YZ$ through $Y$ $O$ be the center of $(XYZ)$ $U = XZ \cap \Gamma_1$ $V = YZ \cap \Gamma_2$ $M_1,M_2$ be the midpoints of $XZ$ and $YZ$ respectively Let $\measuredangle$ and $\angle$ be directed and standard angles respectively. Claim: $XZ=YZ=R_{\Gamma_1}+R_{\Gamma_2}$ ProofSince $XU$ and $YV$ are diameters, then $\angle UBX = 90^{\circ}, \angle VBY = 90^{\circ}$ Therefore, $U,V,B$ are collinear. Since $$\measuredangle ZXY = \measuredangle C_1XB = \measuredangle XBC_1 = \measuredangle C_2BY = \measuredangle BYC_2 = \measuredangle XYZ,$$then $XZ=YZ$ Since $$\measuredangle ZUV = \measuredangle XUB = 90^{\circ} - \measuredangle YXZ = 90^{\circ} - ZYX = \measuredangle BVY =\measuredangle UVZ,$$then $UZ=VZ$ Therefore,$$XZ+YZ=XU+UZ=YZ=XU+VZ+YZ=XU+YV=2R_{\Gamma_1}+2R_{\Gamma_2}$$$$\implies XZ=YZ=R_{\Gamma_1}+R_{\Gamma_2}$$Thus, the Claim is proven ${}_\blacksquare$ Claim: Points $A,C_1,C_2,O,R,Z$ are cyclic ProofClearly, $\measuredangle OM_1Z = \measuredangle OM_2Z = 90^{\circ}$. Therefore, $ZM_1OM_2$ is cyclic Note that $$C_1M_1=XM_1-XC_1=\frac{1}{2} XZ - R_{\Gamma_1} = \frac{1}{2} (R_{\Gamma_1}+R_{\Gamma_2}) - R_{\Gamma_1} = \frac{1}{2}R_{\Gamma_2} - \frac{1}{2}R_{\Gamma_1}$$$$C_2M_2 = YC_2 - YM_2 = R_{\Gamma_2} - frac{1}{2} YZ = R_{\Gamma_2} - frac{1}{2} (R_{\Gamma_1}+R_{\Gamma_2}) = \frac{1}{2}R_{\Gamma_2} - \frac{1}{2}R_{\Gamma_1}$$Therefore, $C_1M_1=C_2M_2$ Since $XZ=YZ$, then $\triangle XYZ$ is isosceles, so by symmetry $OM_1=OM_2$ Since $$C_1M_1=C_2M_2, OM_1=OM_2, \angle OM_1C_1= 90^{\circ} = \angle OM_2C_2,$$then $\triangle OM_1C_1 \cong \triangle OM_2C_2$ Therefore, $$\measuredangle ZC_1O = \measuredangle M_1C_1O = \measuredangle M_2C_2O = \measuredangle ZC_2O,$$so $O$ lies on $(ZC_1C_2)$ Since $$\measuredangle C_1AC_2 = \measuredangle C_2BC_1 = \measuredangle XBC_1 + \measuredangle C_2BY = \measuredangle C_1XB + \measuredangle BYC_2 = \measuredangle XZY = \measuredangle C_1ZC_2,$$then $A$ lies on $(ZC_1C_2)$ as well Similarly, $$\measuredangle C_1AC_2 = \measuredangle C_2BC_1 = \measuredangle PBC_1 + \measuredangle C_2BQ = \measuredangle C_1PB + \measuredangle BQC_2 = \measuredangle PRQ = \measuredangle C_1RC_2,$$so $R$ lies on $(ZC_1C_2)$ which contains $O$ and $Z$ Therefore, points $A,C_1,C_2,O,R,Z$ are cyclic Thus, the Claim is proven ${}_\blacksquare$ Claim: Points $A,P,Q,R$ are cyclic ProofSince $$\measuredangle C_1PA = \measuredangle PAC_1 = 90^{\circ} - \measuredangle ABP = 90^{\circ} - \measuredangle ABQ = \measuredangle C_2QA = \measuredangle QAC_2,$$then $\triangle AC_1P \sim \triangle AC_2Q$ Therefore, $A$ is the center of the spiral similarity sending $PC_1$ to $QC_2$, so it is also the center of the spiral similarity sending $C_1C_2$ to $PQ$ Therefore, $\triangle AC_1C_2 \sim \triangle APQ$, so $\measuredangle PAQ = \measuredangle C_1AC_2 = \measuredangle C_1RC_2 = \measuredangle PRQ$ Therefore, points $A,P,Q,R$ are cyclic Thus, the Claim is proven ${}_\blacksquare$ Claim: $S = PX \cap QY$ ProofBy the Incenter-Excenter Lemma, $S$ is the midpoint of arc $\overarc{PQ}$ of $(PQR)$ Let $S’=PX \cap QY$ It suffices to show that $S’$ is also the midpoint of $\overarc{PQ}$ of $(PQR)$ to get $S=S’$ But note that $A$ is the Miquel point of complete quadrilateral $BXS’Q$, so $A$ is the center of the spiral similarity sending $PQ$ to $XY$ Therefore, $$\measuredangle PS’Q = \measuredangle S’PQ + \measuredangle PQS’ = \measuredangle XPB + \measuredangle BQY$$$$= \measuredangle XAB + \measuredangle BAY = \measuredangle XAY = \measuredangle PAQ = \measuredangle PZQ$$$\implies S’$ lies on $(PQR)$ Since $$\measuredangle C_1XA = \measuredangle XAC_1 = 90^{\circ} - \measuredangle ABX = 90^{\circ} - \measuredangle ABY = \measuredangle C_2YA = \measuredangle YAC_2,$$then $\triangle C_1BX \sim \triangle C_2BY$ Therefore, $$\frac{AX}{AY}=\frac{C_1X}{C_2Y}=\frac{BX}{BY}$$so $AB$ bisects $\angle XAY$ Therefore, $$\measuredangle S’PQ = \measuredangle XPB = \measuredangle XAB = \measuredangle BAY = \measuredangle BQY = \measuredangle PQS’$$so $S’P=S’Q$ Therefore, $S’$ is indeed the midpoint of $\overarc{PQ}$ in $(PQR)$ Thus, the Claim is proven ${}_\blacksquare$ Claim: Points $K,S,X,T,Z$ are cyclic ProofClearly, $K,X,Y,Z$ are cyclic by $\measuredangle KXZ = 90^{\circ} = \measuredangle KYZ$ Therefore, $$\measuredangle XSY = \measuredangle PSQ = \measuredangle PAQ = \measuredangle C_1AC_2 = \measuredangle C_1ZC_2 = \measuredangle XZY$$So $S$ is cyclic with $K,X,Y,Z$ also Thus, the Claim is proven ${}_\blacksquare$ Since $S$ lies on $(XYZ)$, which is fixed, then as $P$ and $Q$ vary, $S$ will trace an arc of the circle $(XYZ)$, and its center $O$ is indeed concyclic with $A,C_1,C_2$ ${}_\blacksquare$