$\left[\begin{array}{l}\text { Law of Cosines } \\ {[\text{ Tangent Circles }}\end{array}\right]$

Two circles with radii $R$ and $r(R>r)$ are externally tangent at point $A$. A line through point $B$, taken on the larger circle, is tangent to the smaller circle at point $C$. Find $B C$, if $A B=a$.
Let $O$ and $Q$ be the centers of the larger and smaller circles, respectively. Since the line connecting the centers of tangent circles passes through their point of tangency, the points $O, Q$, and $A$ lie on the same straight line. Applying the Law of Cosines to the isosceles triangle $A O B$ and to the triangle $O B Q$, we get:

$$
\begin{aligned}
& \cos \angle A O B=\frac{O A^{2}+O B^{2}-A B^{2}}{2 O A O B}=\frac{1 R^{2}+R^{2}-a^{2}}{2 R^{2}}=\frac{2 R^{2}-a^{2}}{2 R^{2}} \\
& B Q 2=O B 2+O Q 2-2 O B \cdot O Q \cos \angle B O Q= \\
& =R 2+(R+r) 2-2 R(R+r) \cdot \frac{2 R^{2}-a^{2}}{2 R^{2}}=\frac{(R+r) a^{2}}{R}+r 2
\end{aligned}
$$

By the Pythagorean theorem,

$$
B C 2=B Q 2-Q C 2=\frac{(R+r) a^{2}}{R}+r 2-r 2=\frac{(R+r) a^{2}}{R} .
$$

Therefore, $B C=a \sqrt{\frac{R+\varphi}{R}}=a \sqrt{1+\frac{\varphi}{R}}$.

## Answer

$a \sqrt{1+\frac{\varphi}{R}}$