3. [7 points] Let $M$ be a figure on the Cartesian plane consisting of all points $(x, y)$ such that there exists a pair of real numbers $a, b$ for which the system of inequalities is satisfied

$$
\left\{\begin{array}{l}
(x-a)^{2}+(y-b)^{2} \leqslant 10 \\
a^{2}+b^{2} \leqslant \min (-6 a-2 b ; 10)
\end{array}\right.
$$

Find the area of the figure $M$.
Answer: $30 \pi-5 \sqrt{3}$.

Solution. The second inequality is equivalent to the system of inequalities

$$
\left\{\begin{array}{l}
a^{2}+b^{2} \leqslant-6 a-2 b \\
a^{2}+b^{2} \leqslant 10
\end{array}\right.
$$

Thus, the original system is equivalent to the following:

$$
\left\{\begin{array} { l } 
{ ( x - a ) ^ { 2 } + ( y - b ) ^ { 2 } \leqslant 1 0 , } \\
{ a ^ { 2 } + b ^ { 2 } \leqslant - 6 a - 2 b , } \\
{ a ^ { 2 } + b ^ { 2 } \leqslant 1 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(a-x)^{2}+(b-y)^{2} \leqslant 10 \\
(a+3)^{2}+(b+1)^{2} \leqslant 10 \\
a^{2}+b^{2} \leqslant 10
\end{array}\right.\right.
$$

The sets of points defined by these inequalities on the plane $(a ; b)$ (where $x$ and $y$ act as parameters) are circles $\omega_{1}, \omega_{2}, \omega_{3}$ with radius $\sqrt{10}$ and centers at $P(x ; y), B(-3 ;-1), A(0 ; 0)$, respectively. The condition of the problem means that the system (36) must have a solution with respect to $(a ; b)$, that is, all three circles must have at least one common point.

Let the circles bounding $\omega_{2}$ and $\omega_{3}$ intersect at points $C$ and $D$ (then triangles $A B C$ and $A B D$ are equilateral). The intersection of circles $\omega_{2}$ and $\omega_{3}$ is a figure $F$, which is the union of two smaller segments of these circles, bounded by the chord $C D$. Then the figure $M$ consists of all possible points $(x ; y)$ that are at a distance of no more than $\sqrt{10}$ from the figure $F$. (This is the union of all circles of radius $\sqrt{10}$, the centers of which belong to the figure $F$.)

Let points $P$ and $Q$ be symmetric to points $A$ and $B$ (respectively) relative to point $C$; points $T$ and $R$ be symmetric to points $A$ and $B$ (respectively) relative to point $D$. It is not difficult to understand that $M$ is the union of four sectors (the central angle of all sectors is less than $180^{\circ}$):

- sector $P A T$ of the circle with center at point $A$ and radius $A P$,
- sector $Q B R$ of the circle with center at point $B$ and radius $B Q$,
- sector $P C Q$ of the circle with center at point $C$ and radius $C P$,
- sector $R D T$ of the circle with center at point $D$ and radius $D T$.

Note that the first two sectors intersect along the rhombus $A C B D$, and there are no other intersections between the sectors. At the same time, the first two sectors are equal to each other, and the last two sectors are also equal to each other. Thus, the area of the figure $M$ is

$S_{M}=S_{P A T}+S_{Q B R}+S_{P C Q}+S_{R D T}-S_{A C B D}=2 \cdot \frac{\pi(2 \sqrt{10})^{2}}{3}+2 \cdot \frac{\pi(\sqrt{10})^{2}}{6}-\frac{\sqrt{3}}{2} \cdot(\sqrt{10})^{2}=30 \pi-5 \sqrt{3}$.