4. Two non-intersecting circles are arranged such that one of their common internal tangents is perpendicular to one of their common external tangents. Find the area of the triangle formed by these tangents and the third common tangent of the circles, if their radii are $r_{1}$ and $r_{2}$.
5.I.4. We will use the following notations and formulas. $S$ - the area of the triangle, $p$ - its semiperimeter, $r$ - the radius of the inscribed circle in the triangle. It is known that $S=p r$. If $r_{a}$ is the radius of the circle touching the side $B C$ of triangle $A B C$ and the extensions of its two other sides, then the formula $S=(p-a) r$ holds.

Assume for definiteness that $r_{1}>r_{2}$. The third common tangent of these circles can be either their external or internal tangent.

Case 1. Suppose the third common tangent is their external tangent. Notations are shown in the figure. Let $a=BC$, $b=AC$ - the legs of the right triangle $A B C$, $c=AB$ - its hypotenuse. Since $r_{1}$ is the radius of the circle inscribed in this triangle, and $r_{2}$ is the radius of its exscribed circle, then

$$
\left\{\begin{array}{l}
a+c-b=\frac{2 S}{r_{1}}, \quad \text { so } \quad b=S\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right) \\
a+b+c=\frac{2 S}{r_{2}},
\end{array}\right.
$$

On the other hand, since this triangle is a right triangle, then $r_{2}=\frac{1}{2}(a+b-c)$ and $r_{1}=\frac{1}{2}(b+c-a)$, from which $b=r_{1}+r_{2}$, therefore

$$
S=\frac{r_{1} r_{2}\left(r_{1}+r_{2}\right)}{r_{1}-r_{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_46613852391e49da1fa9g-089.jpg?height=460&width=1262&top_left_y=1512&top_left_x=387)

Case 2. Suppose the third common tangent is internal. Notations are shown in the figure. In this case, both given circles are exscribed, therefore

$$
\left\{\begin{array}{l}
a+b-c=\frac{2 S}{r_{1}}, \\
a+c-b=\frac{2 S}{r_{2}},
\end{array} \quad \text { from which } \quad a=S\left(\frac{1}{r_{1}}+\frac{1}{r_{2}}\right)\right.
$$

Since $r_{2}=\frac{1}{2}(a+b+c)$ and $r_{1}=\frac{1}{2}(b+c-a)$, then $a=r_{1}-r_{2}$, therefore

$$
S=\frac{r_{1} r_{2}\left(r_{1}-r_{2}\right)}{r_{1}+r_{2}}
$$

Answer: $\frac{r_{1} r_{2}\left(r_{1}+r_{2}\right)}{\left|r_{1}-r_{2}\right|}$ or $\frac{r_{1} r_{2}\left|r_{1}-r_{2}\right|}{r_{1}+r_{2}}$.