7. Let $\omega$ and $\Gamma$ by circles such that $\omega$ is internally tangent to $\Gamma$ at a point $P$. Let $A B$ be a chord of $\Gamma$ tangent to $\omega$ at a point $Q$. Let $R \neq P$ be the second intersection of line $P Q$ with $\Gamma$. If the radius of $\Gamma$ is 17 , the radius of $\omega$ is 7 , and $\frac{A Q}{B Q}=3$, find the circumradius of triangle $A Q R$.
Answer: $\sqrt{170}$
Let $r$ denote the circumradius of triangle $A Q R$. By Archimedes Lemma, $R$ is the midpoint of arc $A B$ of $\Gamma$. Therefore $\angle R A Q=\angle R P B=\angle R P A$ so $\triangle R A Q \sim \triangle R P A$. By looking at the similarity ratio between the two triangles we have
$$
\frac{r}{17}=\frac{A Q}{A P}
$$

Now, let $A P$ intersect $\omega$ again at $X \neq P$. By homothety we have $X Q \| A R$ so
$$
\frac{A X}{A P}=1-\frac{P Q}{P R}=1-\frac{7}{17}=\frac{10}{17}
$$

But we also know
$$
A X \cdot A P=A Q^{2}
$$
so
$$
\frac{10}{17} A P^{2}=A Q^{2}
$$

Thus
$$
\frac{r}{17}=\frac{A Q}{A P}=\sqrt{\frac{10}{17}}
$$
so we compute $r=\sqrt{170}$ as desired.