Assume without loss of generality that the third circle $ \Gamma$ is tangent internally to the first two given ones (the other case can be treated analogously). Denote by $ X^{\prime}$, and by $ Y^{\prime}$ the intersections of the line $ XY$ with the first, and second circle respectively, and let $ O_{1}$, $ O_{2}$ be the centers of these first two given circles $ \gamma_{1}$, and $ \gamma_{2}$; finally let $ Q$ be the center of $ \Gamma$. Since $ \angle{O_{1}XX^{\prime}}=\angle{O_{1}X^{\prime}X}=\angle{QXY}=\angle{QYX}=\angle{O_{2}YY^{\prime}}=\angle{O_{2}Y^{\prime}Y},$ we have that $ O_{1}X \| O_{2}Y^{\prime}$, and $ O_{1}Y \| O_{1}X^{\prime}$. Thus, $ \angle{AXX^{\prime}}=\angle{BY^{\prime}Y}$, and since $ AB$ is tangent to $ \gamma_{2}$, we conclude that $ \angle{AXX^{\prime}}=\angle{BY^{\prime}Y}=\angle{ABY}$. This shows that the points $ A$, $ B$, $ X$, and $ Y$ lie on a same circle.

Let $ \omega_1 ,\omega_2$ denote the two given circles and $ \omega$ denotes the third circle internally/externally tangent to $ \omega_1 ,\omega_2$ at $ X,Y.$ Let $ O$ be the exsimilicenter of $ \omega_1 \sim \omega_2,$ $ X,Y$ are the exsimilicenters/insimilicenters of $ \omega \sim \omega_1$ and $ \omega \sim \omega_2.$ Therefore, by Monge & d'Alembert theorem $ XY$ passes through $ O$ $\Longrightarrow$ $ X,Y$ and $ A,B$ are inverse points under the direct inversion through pole $ O$ that takes $ \omega_1$ and $ \omega_2$ into each other $ \Longrightarrow$ $ A,B,X,Y$ are concyclic $ \Longrightarrow$ $ P \equiv AX \cap BY$ moves on the radical axis of $ \omega_1 ,\omega_2.$

First, let's see that $\{T \} \equiv AX \cap BY$ belongs to the third circle. Proof: if $\omega_1$ and $\omega_2$ are the first two circles and $\Omega$ - the third one, then $X$ is the similicenter (in- or ex-) of $\omega_1$ and $\Omega$ and, obviously, if $\{X, T' \} \equiv AX \cap \Omega$, then the tangent to $\Omega$ at $T'$ is parallel to $AB$, the same for $B$ and $T"$, $T"$ being the intersection of $BY$ with $\Omega$, hence $T" \equiv T' \equiv T$. Next, if $(TZ$ is the direction $(BA$ of this tangent, we have $\widehat{BAT}=\widehat{XTZ}$ $(1)$, but from the circle $\Omega $: $\widehat{XTZ}=\widehat{XYT}$, $(2)$, from $(1)$ and $(2)$ getting that $ABYX$ is cyclic, i.e. $T$ belongs to the radical axis of $\omega_1$ and $\omega_2$. Best regards, sunken rock