4. On the plane $(x ; y)$, plot the set of points whose coordinates satisfy the system $\left\{\begin{array}{l}(|x|+x)^{2}+(|y|-y)^{2} \leq 16, \\ y-3 x \leq 0\end{array}\right.$, and find the area of the resulting figure.
Answer: $\frac{20}{3}+\pi$.

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

1) $x \geq 0, y \geq 0$ (first quadrant). Then $4 x^{2} \leq 16,|x| \leq 2$; in this case, $x$ is positive, so $x \leq 2$.
2) $x<0, y \geq 0$ (second quadrant). Then $0 \leq 16$, the inequality is satisfied by all points in the second quadrant.
3) $x<0, y<0$ (third quadrant). Then $4 y^{2} \leq 16,|y| \leq 2$; in this case, $y$ is negative, so $y \geq-2$.
4) $x \geq 0, y<0$ (fourth 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.

The second inequality defines the set of points located on the line $y=3 x$ 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 fourth quadrant (area $\pi$),
- a triangle with vertices $O, A\left(-\frac{2}{3} ;-2\right), B(0 ;-2)$ (area $\left.\frac{2}{3}\right)$,

"Phystech-2016", mathematics, solution to ticket 16

- a triangle with vertices $O, C(2 ; 0), D(2 ; 6)$ (area 6). Therefore, the area of the set is $\frac{20}{3}+\pi$.