Let the intersection point of $\Omega _1$ and $\Omega _2$ be $Q \neq P$. Since $\angle X_1 T X_2 = 180 - \angle T X_1 X_2 - \angle T X_2 X_1 = 180 - \angle X_1 Q P - \angle X_2 Q P = 180 - \angle X_1 Q X_2$, $\angle X_1 Q X_2 + \angle X_1 T X_2 = 180$, therefore $(X_1, T, X_2, Q)$ are cyclic. Let the circle that passes through $X_1, T, X_2, Q$ be $\omega$. Let $S$ be the intersection point of $\omega$ and line $QP$. Let the perpendicular bisector of $X_1 X_2$ be line $m$. Since $\angle X_1 Q S = \angle X_1 Q P = \angle X_2 X_1 T = \angle X_2 Q T$, points $S$ and $T$ are symmetrical with respect to line $m$. also, since $X_1 P=X_2 R$, points $P$ and $R$ are symmetrical with respect to line $m$. Therefore, the two lines $SP$ and $TR$ are symmetrical with respect to line $m$. Claim) line $m$ passes through a fixed point, independent of the choice of $l$. proof) Let the midpoint of $X_1 X_2$ be $N$. It is well known that the trace of $N$ is a circle that passes through $P$ and $Q$. Let the trace of $N$ be $\Gamma$. Let the antipode of $P$ in $\Gamma$ be $P^*$. Then, $\angle P M P^* = 90$, therefore $M P^*$ is the perpendicular bisector of line $X_1 X_2$. Since $P^*$ is fixed, Claim) is proved. Let $\gamma$ be a circle with center $P^*$ and is tangent to $QP$. Since $PS$ and $RT$ is symmetrical WRT $m$, and the center of $\gamma$ is on $m$, $\gamma$ is also tangent to $RT$. Also, $\gamma$ is fixed since the center $P*$ is fixed, and the radius, which is the distance of $PQ$ and $P^*$, is fixed. Therefore, $RT$ is tangent to the fixed circle $\gamma$ independent of the choice of $l$.

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