4. Let $A, B, C$ be the centers of the given circles, $A B=2, A C=2, B C=3$ (see Fig. 1).

Denote by $a, b$, and $c$ the radii of the circles centered at points $A, B$, and $C$ respectively, and find their values by solving the system of equations. We get:

$$
\left\{\begin{array}{l}
a + b = 2 \\
a + c = 2 \\
b + c = 3
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a=\frac{1}{2} \\
b=\frac{3}{2} \\
c=\frac{3}{2}
\end{array}\right.\right.
$$

Let $S$ and $L$ be the centers of the small and large circles, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_06_56ffd9f61e4fedda1946g-2.jpg?height=617&width=625&top_left_y=157&top_left_x=1178)

Fig. 1

Denote by $r$ and $R$ the radii of the small and large circles that are tangent to the three given circles, respectively.

Let $\angle A S B=2 \gamma, \angle A S C=2 \gamma$, and $\angle B S C=2 \alpha$. Then, by the cosine rule for triangle $A S B$, we get:

$$
\cos 2 \gamma=\frac{(a+r)^{2}+(b+r)^{2}-(a+b)^{2}}{2(a+r)(b+r)}=\frac{r^{2}+(a+b) r-a b}{(a+r)(b+r)}
$$

From this, using the double-angle formulas, we find:

$$
\cos ^{2} \gamma=\frac{r(a+b+r)}{(a+r)(b+r)}, \sin ^{2} \gamma=\frac{a b}{(a+r)(b+r)}
$$

Similarly, for triangle $B S C$, we get:

$$
\sin ^{2} \alpha=\frac{b^{2}}{(b+r)^{2}}
$$

Since $\alpha+2 \gamma=\pi$, then $\sin ^{2} \alpha=\sin ^{2} 2 \gamma=4 \sin ^{2} \gamma \cos ^{2} \gamma$.

From this, we have:

$$
\begin{gathered}
\frac{b^{2}}{(b+r)^{2}}=4 \cdot \frac{a b}{(a+r)(b+r)} \cdot \frac{r(a+b+r)}{(a+r)(b+r)} \\
b^{2}(a+r)^{2}=4 a b(a+b+r) \\
(b-4 a) r^{2}+\left(2 a b^{2}-4 a^{2}-4 a b\right) r+a^{2} b=0
\end{gathered}
$$

Substituting the found values of the radii of the original circles, we get the equation for determining the value of $r$:

$$
\begin{aligned}
& r^{2}+5 r-\frac{3}{4}=0 \\
& r=\frac{1}{2}(2 \sqrt{7}-5)
\end{aligned}
$$

Note that to find the value of the radius of the large circle $R$, in all the equations for finding the radius of the small circle, we can replace $r$ with $-R$. Then we get the equation:

$$
\begin{aligned}
& R^{2}-5 R-\frac{3}{4}=0 \\
& R=\frac{1}{2}(2 \sqrt{7}+5)
\end{aligned}
$$

We find the area of the desired figure, we get: $S=\pi R^{2}-\pi r^{2}=10 \sqrt{7} \pi$.

Note. The radii of the original circles are found - 1 point. The radius of one of the circles is found - 4 points.
Answer: $10 \sqrt{7} \pi$