The radii of circles $S_{1}$ and $S_{2}$, touching at point $A$, are $R$ and $r(R>r)$. Find the length of the tangent line drawn to circle $S_{2}$ from point $B$, lying on circle $S_{1}$, if it is known that $A B=a$. (Consider the cases of internal and external tangency).

#
Apply the theorem of the tangent and secant.

## Solution

Consider the case of external tangency. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$; $X$ be the point of intersection of line $A B$ with circle $S_{2}$, different from $A$; $B M$ be the tangent to circle $S_{1}$ ($M$ is the point of tangency).

Isosceles triangles $X O_{2} A$ and $B O_{1} A$ are similar. Therefore,

$$
A X=\frac{r}{R} \cdot A B=\frac{a r}{R}
$$

By the theorem of the tangent and secant,

$$
B M^{2}=B X \cdot B A=(B A+A X) B A=\left(a+\frac{a r}{R}\right) a=a^{2}\left(1+\frac{r}{R}\right)
$$

Thus, $B M=a \sqrt{1+\frac{r}{R}}$.

In the case of internal tangency, similarly, we obtain that

$$
B M=a \sqrt{1-\frac{r}{R}}
$$

Consider the case of external tangency. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$; $X$ be the point of intersection of line $A B$ with circle $S_{2}$, different from $A$; $B M$ be the tangent to circle $S_{1}$ ($M$ is the point of tangency).

Isosceles triangles $X O_{2} A$ and $B O_{1} A$ are similar. Therefore,

$$
A X=\frac{r}{R} \cdot A B=\frac{a r}{R}
$$

By the theorem of the tangent and secant,

$$
B M^{2}=B X \cdot B A=(B A+A X) B A=\left(a+\frac{a r}{R}\right) a=a^{2}\left(1+\frac{r}{R}\right)
$$

Thus, $B M=a \sqrt{1+\frac{r}{R}}$.

In the case of internal tangency, similarly, we obtain that

$$
B M=a \sqrt{1-\frac{r}{R}}
$$

Consider the case of external tangency. Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$; $X$ be the point of intersection of line $A B$ with circle $S_{2}$, different from $A$; $B M$ be the tangent to circle $S_{1}$ ($M$ is the point of tangency).

Isosceles triangles $X O_{2} A$ and $B O_{1} A$ are similar. Therefore,

$$
A X=\frac{r}{R} \cdot A B=\frac{a r}{R}
$$

By the theorem of the tangent and secant,

$$
B M^{2}=B X \cdot B A=(B A+A X) B A=\left(a+\frac{a r}{R}\right) a=a^{2}\left(1+\frac{r}{R}\right)
$$

Thus, $B M=a \sqrt{1+\frac{r}{R}}$.

In the case of internal tangency, similarly, we obtain that

$$
B M=a \sqrt{1-\frac{r}{R}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_7b30f89ee421e63350f8g-09.jpg?height=440&width=672&top_left_y=169&top_left_x=699)

## Answer

$a \sqrt{1 \pm \frac{r}{R}}$.