2. Find the area of the figure, the coordinates $(x ; y)$ of the points of which satisfy the system of inequalities

$$
\left\{\begin{array}{c}
|x+1|-|y+3| \geq 0 \\
\log _{4}\left(x^{2}+y^{2}+6 y+2 x-26\right) \leq 3
\end{array}\right.
$$
Solution. $\left\{\begin{array}{c}|x+1|-|y+3| \geq 0 \\ \log _{4}\left((x+1)^{2}+(y+3)^{2}-36\right) \leq 3\end{array}\right.$

Substitution: $\mathrm{u}=\mathrm{x}+1, \mathrm{v}=\mathrm{y}+3 /$

We have: $\left\{\begin{array}{c}|u|-|v| \geq 0 \\ 36 \geq v^{2}+u^{2} \leq 100 .\end{array}\right.$

The area of the figure is $S=\frac{\pi(100-36)}{2}=32 \pi$.

Answer: $32 \pi$.