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

Two circles of radii $r$ and $3r$ are externally tangent. Find the area of the figure enclosed between the circles and their common external tangent.
Let $O_1$ and $O_2$ be the centers of circles with radii $r$ and $3r$ respectively, $M$ be the point of tangency of the circles, and $AB$ be the common external tangent of these circles (point $A$ lies on the first circle, $B$ on the second). Drop a perpendicular $O_1F$ from the center of the first circle to the radius $O_2B$ of the second circle. In the right triangle $O_1O_2F$, it is known that

$$
O_1O_2 = O_1M + O_2M = r + 3r = 4r, \quad O_2F = O_2B - BF = O_2F - O_1A = 3r - r = 2r,
$$

therefore

$$
\angle FO_2O_1 = 60^\circ, \quad \angle AO_1O_2 = 120^\circ, \quad AB = O_1F = 2r\sqrt{3}
$$

Let $S$ be the area of the desired figure (the curvilinear triangle $AMB$), $S_1$ and $S_2$ be the areas of the sectors $AO_1M$ and $BO_2M$, and $S_3$ be the area of the right trapezoid $ABO_2O_1$. Then

$$
S_1 = \frac{1}{3} \pi r^2, \quad S_2 = \frac{1}{6} \pi (3r)^2 = \frac{3}{2} \pi r^2, \quad S_3 = \frac{1}{2} (r + 3r) \cdot 2r\sqrt{3} = 4r^2\sqrt{3}
$$

Therefore,

$$
S = S_3 - S_2 - S_1 = 4r^2\sqrt{3} - \frac{3}{2} \pi r^2 - \frac{1}{3} \pi r^2 = \left(4\sqrt{3} - \frac{11}{6} \pi\right) r^2
$$

## Answer

$\left(4\sqrt{3} - \frac{11}{6} \pi\right) r^2$.

Two circles touch each other externally. Their radii are in the ratio 3:1, and the length of their common external tangent is $6\sqrt{3}$. Find the perimeter of the figure formed by the external tangents and the external parts of the circles.

## Solution

Let $O_1$ and $O_2$ be the centers of circles with radii $r$ and $3r$ respectively, $M$ be the point of tangency of the circles, and $AB$ be the common external tangent of these circles (point $A$ lies on the first circle, $B$ on the second). Drop a perpendicular $O_1F$ from the center of the first circle to the radius $O_2B$ of the second circle. In the right triangle $O_1O_2F$, it is known that

$$
O_1O_2 = O_1M + O_2M = r + 3r = 4r, \quad O_2F = O_2B - BF = O_2F - O_1A = 3r - r = 2r,
$$

therefore

$$
\angle FO_2O_1 = 60^\circ, \quad \angle AO_1O_2 = 120^\circ, \quad AB = O_1F = 2r\sqrt{3} = 6\sqrt{3},
$$

from which $r = 3, 3r = 9$. Let $P$ be the perimeter of the desired figure, $l_1$ and $l_2$ be the lengths of the external parts of the first and second circles respectively.

$$
l_1 = \frac{1}{3} \cdot 2\pi r = \frac{1}{3} \cdot 2\pi \cdot 3 = 2\pi, \quad l_2 = \frac{2}{3} \cdot 2\pi \cdot 3r = \frac{2}{3} \cdot 2\pi \cdot 9 = 12\pi
$$

Therefore,

$$
P = 2AB + l_1 + l_2 = 12\sqrt{3} + 14\pi
$$

## Answer

$14\pi + 12\sqrt{3}$.