## Task 3 - 200923

Given two circles $k_{1}$ and $k_{2}$ with radii $r_{1}$ and $r_{2}$ respectively, where $r_{1}>r_{2}$.

Furthermore, it is assumed that both circles touch each other externally, meaning they have exactly one common internal tangent. This internal tangent intersects one of the common external tangents of both circles at $P$ and the other common tangent at $Q$.

Determine, under these conditions, the length $P Q$ from $r_{1}$ and $r_{2}$!
Let $S$ be the intersection of the two external tangents, $M_{1}$ the center of $k_{1}$, $M_{2}$ the center of $k_{2}$, $B$ their point of tangency, $B_{1}$ the point of tangency of $k_{1}$ with the tangent $t$ on which $P$ lies, and $B_{2}$ the point of tangency of $k_{2}$ with the same tangent $t$.

Then $M_{1} B_{1}$ and $M_{2} B_{2}$ are parallel, as they are both perpendicular to the tangent $t$ as radii of tangency. By the intercept theorem, we have $\frac{\left|S M_{1}\right|}{\left|S M_{2}\right|}=\frac{\left|M_{1} B_{1}\right|}{\left|M_{2} B_{2}\right|}=\frac{r_{1}}{r_{2}}$.

In particular, $\left|S M_{1}\right|>\left|S M_{2}\right|$, so, since $S, M_{2}, B$ and $M_{2}$ lie on a straight line due to symmetry, $\left|S M_{1}\right|=\left|S M_{2}\right|+\left|M_{2} B\right|+\left|B M_{1}\right|=\left|S M_{2}\right|+r_{1}+r_{2}$. Substituting this into the ratio equation obtained, we get

$$
\begin{gathered}
\frac{\left|S M_{2}\right|+r_{1}+r_{2}}{\left|S M_{2}\right|}=1+\frac{r_{1}+r_{2}}{\left|S M_{2}\right|}=\frac{r_{1}}{r_{2}} \quad \text { so } \\
\left|S M_{2}\right|=r_{2} \cdot \frac{r_{1}+r_{2}}{r_{1}-r_{2}} \quad \text { and } \quad\left|S M_{1}\right|=r_{1} \cdot \frac{r_{1}+r_{2}}{r_{1}-r_{2}}
\end{gathered}
$$

$\triangle M_{1} S B_{1}$ is a right triangle at $B_{1}$, so by the Pythagorean theorem,

$$
\begin{gathered}
\left|S B_{1}\right|=\sqrt{\left|S M_{1}\right|^{2}-\left|M_{1} B_{1}\right|^{2}}=\sqrt{r_{1}^{2} \cdot \frac{\left(r_{1}+r_{2}\right)^{2}}{\left.\left(r_{1}-r_{2}\right)^{2}\right)}-r_{1}^{2}}=r_{1} \cdot \sqrt{\frac{\left(r_{1}+r_{2}\right)^{2}-\left(r_{1}-r_{2}\right)^{2}}{\left(r_{1}-r_{2}\right)^{2}}} \\
=\frac{r_{1}}{r_{1}-r_{2}} \cdot \sqrt{4 r_{1} r_{2}}=r_{1} \cdot \frac{2 \sqrt{r_{1} r_{2}}}{r_{1}-r_{2}}
\end{gathered}
$$

and similarly $\left|S B_{2}\right|=r_{2} \cdot \frac{2 \sqrt{r_{1} r_{2}}}{r_{1}-r_{2}}$. Therefore,

$$
\left|B_{1} B_{2}\right|=\left|S B_{1}\right|-\left|S B_{2}\right|=\frac{2 \sqrt{r_{1} r_{2}}}{r_{1}-r_{2}} \cdot\left(r_{1}-r_{2}\right)=2 \sqrt{r_{1} r_{2}}
$$

The point $P$ lies on two tangents to $k_{1}$, so the tangent segments $\left|P B_{1}\right|$ and $|P B|$ are of equal length. For the same reason (tangents to $k_{2}$), $|P B|=\left|P B_{2}\right|$ and thus $P$ is the midpoint of the segment $B_{1} B_{2}$. In particular, $|P B|=\frac{1}{2}\left|B_{1} B_{2}\right|=\sqrt{r_{1} r_{2}}$.

By symmetry, $|Q B|=|P B|$, and since $Q, B$ and $P$ lie on a straight line by assumption,

$$
|P Q|=2|P B|=2 \sqrt{r_{1}} r_{2}
$$