[ Common tangent to two circles [ Trigonometric ratios in a right triangle ]

On the plane, there are two circles with radii 12 and 7, centered at points $O_{1}$ and $O_{2}$, touching a certain line at points $M_{1}$ and $M_{2}$ and lying on the same side of this line. The ratio of the length of the segment $M_{1} M_{2}$ to the length of the segment $O_{1} O_{2}$ is $\frac{2 \sqrt{5}}{5}$. Find $M_{1} M_{2}$.
Consider the right triangle $O_{1} P O_{2}$, where $P$ is the projection of point $O_{2}$ onto $O_{1} M_{1}$.

## Solution

Let $P$ be the projection of point $O_{2}$ onto the line $O_{1} M_{1}$. Denote $\angle O_{1} O_{2} P=\alpha$. Then

$$
\begin{gathered}
\cos \alpha=\frac{O_{2} P}{O_{1} O_{2}}=\frac{M_{1} M_{2}}{O_{1} O_{2}}=\frac{2 \sqrt{5}}{5} \\
\sin \alpha=\frac{\sqrt{5}}{5}, \operatorname{ctg} \alpha=\frac{\cos \alpha}{\sin \alpha}=2
\end{gathered}
$$

In the right triangle $O_{1} \mathrm{PO}_{2}$:

$$
\begin{gathered}
O_{1} P=O_{1} M_{1}-P M_{1}=O_{1} M_{1}-O_{2} M_{2}=12-7=5 \\
M_{1} M_{2}=O_{2} P=O_{1} P \cdot \operatorname{ctg} \alpha=5 \cdot 2=10
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-03.jpg?height=401&width=1056&top_left_y=857&top_left_x=501)

Answer

10.