[ Tangents touching circles $\quad$]

A circle of radius 2 touches another circle externally at point A. The common tangent to both circles, drawn through point $A$, intersects with another common tangent at point $B$. Find the radius of the second circle if $A B=4$.
If $r$ and $R$ are the radii of the circles, then $A B=\sqrt{r R}$.

## Solution

First method.

Let $r$ and $R$ be the radii of the circles ($r=2$), and $C$ and $D$ be the points of tangency of the circles with the second (external) tangent. Then

$$
B C=A B=B D=4
$$

Since $C D=2 \sqrt{r R}$, we have $\sqrt{2 R}=4$. From this, we find that $R=8$.

## Second method.

If $O_{1}$ and $O_{2}$ are the centers of the circles, then $B A$ is the altitude of the right triangle $O_{1} B O_{2}$, dropped from the vertex of the right angle to the hypotenuse $O_{1} O_{2}$. Therefore,

$$
O_{1} A \cdot O_{2} A=A B^{2}=16
$$

From this, we find that $O_{2} A=8$.

## First method.

Let $r$ and $R$ be the radii of the circles ($r=2$), and $C$ and $D$ be the points of tangency of the circles with the second (external) tangent. Then

$$
B C=A B=B D=4
$$

Since $C D=2 \sqrt{r R}$, we have $\sqrt{2 R}=4$. From this, we find that $R=8$.

## Second method.

If $O_{1}$ and $O_{2}$ are the centers of the circles, then $B A$ is the altitude of the right triangle $O_{1} B O_{2}$, dropped from the vertex of the right angle to the hypotenuse $O_{1} O_{2}$. Therefore,

$$
O_{1} A \cdot O_{2} A=A B^{2}=16
$$

From this, we find that $O_{2} A=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-13.jpg?height=420&width=594&top_left_y=59&top_left_x=732)

## Answer

8.