Problem 86. Find the radius of the circle that touches two given circles of radii $R_{1}$ and $R_{2}$ and their common external tangent.
Solution. In the notation of Fig. 87, where $x$ is the radius of the desired circle, we write: $A C + C B = A B$, or, according to Archimedes' problem:

66

$$
2 \sqrt{R_{1} x} + 2 \sqrt{x R_{2}} = 2 \sqrt{R_{1} R_{2}}
$$

From this, it is not difficult to find $x$.

$$
x = \frac{R_{1} R_{2}}{\left(\sqrt{R_{1}} + \sqrt{R_{2}}\right)^{2}}
$$

It might seem that the problem is solved, but it is not advisable to rush to the next problem, as the solution is only half complete!
![](https://cdn.mathpix.com/cropped/2024_05_21_c833ac06d3d78911cbb1g-068.jpg?height=1028&width=1386&top_left_y=360&top_left_x=336)

In Fig. 88, the second variant of constructing the desired circle is shown. In this case, the solution is analogous:

$$
2 \sqrt{R_{1} x} = 2 \sqrt{R_{1} R_{2}} + 2 \sqrt{x R_{2}}
$$

from which:

$$
x = \frac{R_{1} R_{2}}{\left(\sqrt{R_{1}} - \sqrt{R_{2}}\right)^{2}}
$$

Thus,

answer: $x = \frac{R_{1} R_{2}}{\left(\sqrt{R_{1}} \pm \sqrt{R_{2}}\right)^{2}}$