By Archimedes lemma, we have $M, Q, V$ are collinear and $M, P, U$ are collinear as well. Note that inversion at $M$ with radius $MX$ shows that $PQUV$ is cyclic. Thus, we have $MP \cdot MU=MQ \cdot MV$, so $M$ lies on the radical axis of $(PUZ)$ and $(QVZ)$, thus $T$ must lie on the line $MZ$. Thus, we have $MZ \cdot MT=MQ \cdot MV=MN^{2}$, which implies triangles $MZN$ and $MNT$ are similar. Thus, we have $NT \perp MN$. However, since the line through $O_{1}$ and $O_{2}$ passes through $N$ and is perpendicular to $MN$, we have $T$ lies on line $O_{1}O_{2}$. Additionally, since $MZ \cdot MT=MN^{2}=MX^{2}$, inversion at $M$ with radius $MX$ swaps $Z$ and $T$, and since $(MXY)$ maps to line $XY$, this means $T$ also lies on $XY$. Therefore, $T$ is the intersection of $PQ$ and $O_{1}O_{2}$, and thus by Monge's Theorem, we must have $T$ lies on $UV$. Now, to finish, we will consider triangle $OUV$. Since $O_{1}O_{2}T$ is a line that cuts this triangle, by Menelaus, we have $\frac{OO_{1}}{O_{1}U} \cdot \frac{UT}{VT} \cdot \frac{VO_{2}}{O_{2}O}=1$. Using the values of the radii, this simplifies to $\frac{R-r_{1}}{r_{1}} \cdot \frac{UT}{VT} \cdot \frac{r_{2}}{R-r_{2}}=1 \Longrightarrow \frac{UT}{VT}=\frac{r_{1}\left(R-r_{2}\right)}{r_{2}\left(R-r_{1}\right)}$. Now, note that $TU \cdot TV=TP \cdot TQ=\frac{4r_{1}^{2}r_{2}^{2}}{\left(r_{1}-r_{2}\right)^{2}}$. Now, let $TU=r_{1}\left(R-r_{2}\right)k$ and $TU=r_{2}\left(R-r_{1}\right)k$. Plugging this into the above equation gives $r_{1}r_{2}\left(R-r_{1}\right)\left(R-r_{2}\right)k^{2}=\frac{4\left(r_{1}r_{2}\right)^{2}}{\left(r_{1}-r_{2}\right)^{2}}$. Solving gives $k=\frac{2\sqrt{r_{1}r_{2}}}{\left|r_{1}-r_{2}\right|\sqrt{\left(R-r_{1}\right)\left(R-r_{2}\right)}}$. To finish, note that $TU+TV=k\left(Rr_{1}+Rr_{2}-2r_{1}r_{2}\right)=\frac{2\left(Rr_{1}+Rr_{2}-2r_{1}r_{2}\right)\sqrt{r_{1}r_{2}}}{\left|r_{1}-r_{2}\right|\sqrt{\left(R-r_{1}\right)\left(R-r_{2}\right)}}$.