4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}(|x|-x)^{2}+(|y|-y)^{2} \leq 16, \\ 2 y+x \leq 0\end{array}\right.$, and find the area of the resulting figure.
Answer: $5+\pi$.

Solution. Consider the first inequality. There are four possible cases.

1) $x \geq 0, y \geq 0$ (first quadrant). Then $0 \leq 16$, the inequality is satisfied by all points in the first quadrant. In this case, $x$ is positive, so $x \leq 2$.
2) $x<0, y \geq 0$ (second quadrant). Then $4 x^{2} \leq 16,|x| \leq 2$; in this case, $x$ is negative, so $x \geq-2$.
3) $x<0, y<0$ (third quadrant). Then $4 x^{2}+4 y^{2} \leq 16, x^{2}+y^{2} \leq 2^{2}$. We get the points lying on the circle centered at $O(0 ; 0)$ with radius 2 or inside it.
4) $x \geq 0, y<0$ (fourth quadrant). Then $4 y^{2} \leq 16,|y| \leq 2$; in this case, $y$ is negative, so $y \geq-2$.

The second inequality defines the set of points located on the line $y=-\frac{x}{2}$ or below it.

The set of points satisfying the system of inequalities is the union of

- a quarter of a circle centered at $O$ with radius 2, lying in the third quadrant (area $\pi$),
- a triangle with vertices $O, A(-2 ; 1), B(-2 ; 0)$ (area 1),
- a triangle with vertices $O, C(4 ;-2), D(0 ;-2)$ (area 4).

“Phystech-2016”, mathematics, solution to ticket 14

Thus, the area of the set is $5+\pi$.