Let $r_1$ be the radius of $C_1$ and $r_2$ the radius of $C_2$. $N$ is an homothetic center. $P$ and $Q$ are homologous, so are $M$ and $S$ and the centers of both circunference. $\frac{NS}{NM}=\frac{r_2}{r_1}=3$ and $NS=NM+2r_1$ therefore, $NM=r_1$ so $NO_2=6r_1$ and $S$ is its midpoint. $30=\angle QNS=\angle NQS=\angle QTS=\angle TQR$.

Math_CYCR wrote: Let $C_1$ and $C_2$ be two circumferences externally tangents at $S$ such that the radius of $C_2$ is the triple of the radius of $C_1$. Let a line be tangent to $C_1$ at $P \neq S$ and to $C_2$ at $Q \neq S$. Let $T$ be a point on $C_2$ such that $QT$ is diameter of $C_2$. Let the angle bisector of $\angle SQT$ meet $ST$ at $R$. Prove that $QR=RT$ Let $OE \bot KQ \Leftrightarrow KE = 2r = \frac{{OK}}{2} \Leftrightarrow K\widehat OE = {30^0}$. The rest is easy. All angles are in the figure.