2. Plot on the plane $(x ; y)$ the set of points satisfying the equation $\left|9-x^{2}-y^{2}-2 y\right|+|-2 y|=9-x^{2}-y^{2}-4 y$, and find the area of the resulting figure.
Answer: $S=10 \pi+3-10 \operatorname{arctg} 3$.

Solution. Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right side). Therefore, the first equation is equivalent to the system of inequalities

$$
\left\{\begin{array} { l } 
{ 9 - x ^ { 2 } - y ^ { 2 } - 2 y \geq 0 , } \\
{ - 2 y \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(y+1)^{2}+x^{2} \leq 10 \\
y \leq 0
\end{array}\right.\right.
$$

The first inequality defines a circle of radius $\sqrt{10}$ centered at $(0 ;-1)$, and the entire system defines the part of this circle lying in the half-plane $y \leq 0$. The area of this segment is the difference between the area of the circle and the area of the segment of this circle located in the half-plane $y \geq 0$. Since the central angle of this segment is $2 \operatorname{arctg} 3$, we obtain that its area $S$ is $10 \pi+3-10 \operatorname{arctg} 3$.