5. Let $A$ and $E$ be the centers of circles $\omega_{1}$ and $\omega_{2}$ with radii $r$ and $R$, respectively $(r<R)$, and by the problem's condition, $A \in \omega_{2}$, while $B$ and $C$ are the points of intersection of these circles. Note that their common chord $BC$ is perpendicular to the segment $AE$. We have $AB = AC = r, \quad EB = EC = EA = R$. Let $\square BAE = \square CAE = \alpha, \quad \square BEA = \square CEA = \gamma$.

![](https://cdn.mathpix.com/cropped/2024_05_06_db9ce7a1cbcec365545eg-06.jpg?height=402&width=545&top_left_y=183&top_left_x=479)

Write the formula for calculating the area of the union of circles $\Omega_{1}$ and $\Omega_{2}$, bounded by circles $\omega_{1}$ and $\omega_{2}$:

$$
S_{\Omega_{1} \cup \Omega_{2}} = S_{\Omega_{1}} + S_{\Omega_{2}} - S_{\Omega_{1} \cap \Omega_{2}},
$$

where $S_{\Omega_{1}} = \pi r^{2}, S_{\Omega_{2}} = \pi R^{2}$ are the areas of circles $\Omega_{1}$ and $\Omega_{2}$, and $S_{\Omega_{1} \cap \Omega_{2}}$ is the area of their intersection.

Find $S_{\Omega_{1} \cap \Omega_{2}}$. The intersection $\Omega_{1} \cap \Omega_{2}$ consists of two segments $BDC \in \Omega_{1}$ and $BAC \in \Omega_{2}$. Denote the areas of these segments by $S_{1}$ and $S_{2}$, respectively.

The area $S_{1}$ of segment $BDC$ is the difference between the area of sector $BDCA$ and the area of $\square ABC$:

$$
S_{1} = \frac{(2\alpha) r^{2}}{2} - \frac{1}{2} r^{2} \sin 2\alpha
$$

Since $\square ABE$ is isosceles $(AE = BE)$, we have $\cos \alpha = \frac{r}{2R}$. Then

$$
\begin{gathered}
\alpha = \arccos \frac{r}{2R}, \sin \alpha = \sqrt{1 - \left(\frac{r}{2R}\right)^{2}} \\
\sin 2\alpha = 2 \sin \alpha \cos \alpha = \frac{r}{2R} \cdot \sqrt{1 - \left(\frac{r}{2R}\right)^{2}}
\end{gathered}
$$

By the problem's condition, $r = 1, \quad R = 2$. Substituting these values into the above formulas: $\quad \cos \alpha = \frac{1}{4}, \quad \alpha = \arccos \frac{1}{4}$, $\sin \alpha = \frac{\sqrt{15}}{4}, \sin 2\alpha = \frac{\sqrt{15}}{8}$. Then

$$
S_{1} = \arccos \frac{1}{4} - \frac{\sqrt{15}}{16}
$$

The area $S_{2}$ of segment $BAC$ is the difference between the area of sector $BACE$ and the area of $\square CBE$:

$$
S_{2} = \frac{(2\gamma) R^{2}}{2} - \frac{1}{2} R^{2} \sin 2\gamma
$$

Since $\square ABE$ is isosceles $(AE = BE)$, we have $\gamma = \pi - 2\alpha$. Find

$$
\begin{gathered}
\cos \gamma = \cos (\pi - 2\alpha) = -\cos 2\alpha = \sin^{2} \alpha - \cos^{2} \alpha = \frac{15}{16} - \frac{1}{16} = \frac{7}{8} \\
\sin \gamma = \sqrt{1 - \frac{49}{64}} = \frac{\sqrt{15}}{8}, \sin 2\gamma = \frac{7\sqrt{15}}{32}
\end{gathered}
$$

Then

$$
S_{2} = 4\left(\pi - 2 \arccos \frac{1}{4}\right) - \frac{7\sqrt{15}}{16}
$$

Calculate the area of the intersection of circles $\Omega_{1}$ and $\Omega_{2}$:

$$
S_{\Omega_{1} \cap \Omega_{2}} = S_{1} + S_{2} = 4\pi - 7 \arccos \frac{1}{4} - \frac{\sqrt{15}}{2}
$$

Substituting the obtained result into the formula for calculating the area of the union of circles $\Omega_{1}$ and $\Omega_{2}$, we find:

$$
S_{\Omega_{1} \cup \Omega_{2}} = \pi + 4\pi - \left(4\pi - 7 \arccos \frac{1}{4} - \frac{\sqrt{15}}{2}\right) = \pi + 7 \arccos \frac{1}{4} + \frac{\sqrt{15}}{2}.
$$
Answer: $S=\pi+7 \arccos \frac{1}{4}+\frac{\sqrt{15}}{2}$.