8,9

The ratio of the radii of circles $S 1$ and $S 2$, touching at point $B$, is $k(k>1)$. A line is drawn from point $A$, lying on circle $S 1$, tangent to circle $S 2$ at point $C$. Find $A C$, given that the chord cut by circle $S 2$ on line $A B$ is equal to $b$.
Consider the case of external tangency. Let $O_1$ and $O_2$ be the centers of circles $S_1$ and $S_2$ with radii $R$ and $r$ respectively, $\frac{R}{\tau}=k>1$; $X$ is the point of intersection of line $AB$ with circle $S_2$, different from $A$, and $AX=b$.

Isosceles triangles $XO_2A$ and $BO_1A$ are similar. Therefore,

$$
AB=\frac{\mathscr{R}}{\tau} \cdot AX=\frac{b \tau}{R}
$$

By the tangent-secant theorem,

$$
BC^2=BX \cdot BA=(BA + AX) \cdot BA = \left(b + \frac{b \tau}{\pi}\right) \frac{b \tau}{T} = b^2 (1 + k) k
$$

Thus, $BC = b \sqrt{k + k^2}$.

In the case of internal tangency, similarly, we obtain $BC = b \sqrt{k^2 - k}$.

## Answer

$b \sqrt{k^2 \pm k}$.