6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities

$$
\left\{\begin{array}{l}
y-x \geqslant|x+y| \\
\frac{x^{2}+8 x+y^{2}+6 y}{2 y-x-8} \leqslant 0
\end{array}\right.
$$

Sketch the figure $M$ and find its area.
Answer: 8.

Solution. The first inequality is equivalent to the system ${ }^{1}$ $\left\{\begin{array}{l}x+y \leqslant y-x, \\ x+y \geqslant x-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leqslant 0, \\ y \geqslant 0 .\end{array}\right.\right.$

Consider the second inequality. It can be written as $\frac{(x+4)^{2}+(y+3)^{2}-25}{2 y-x-8} \leqslant 0$. The numerator of the fraction on the left side of the inequality is zero on the circle of radius 5 centered at the point $Q(-4 ;-3)$ (let's call it $\omega$). The denominator of the fraction is zero on the line $y=4+\frac{x}{2}$ (let's call it $\ell$). The points of intersection of the circle and the line are determined from the system of equations

$$
\left\{\begin{array}{l}
x = 2 y - 8 \\
x ^ { 2 } + 8 x + y ^ { 2 } + 6 y = 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x = 2 y - 8 \\
( 2 y - 8 ) ^ { 2 } + 8 ( 2 y - 8 ) + y ^ { 2 } + 6 y = 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=2 y-8 \\
y^{2}-2 y=0
\end{array}\right.\right.\right.
$$

from which we get two points $A(-4 ; 2)$ and $B(-8 ; 0)$. Also denote the origin by $O$, and the intersection of the line $\ell$ with the $O y$ axis by $C$ (it is not difficult to determine that the coordinates of point $C$ are $(0 ; 4)$).

The inequality is satisfied:

- at all points of the circle $\omega$ except points $A$ and $B$ (then the numerator of the fraction is zero);
- inside the circle $\omega$ at points located above the line $\ell$ (the numerator is negative, and the denominator is positive);
- outside the circle $\omega$ at points located below the line $\ell$ (the numerator is positive, and the denominator is negative).

Describe the set of points $M$ that satisfy the original system of inequalities. It consists of the segment of the circle bounded by the chord $A B$ and located above this chord, as well as the curvilinear triangle $A O C$, the boundaries of which are the arc $A O$ of the circle $\omega$ and the segments $A C$ and $C O$ (while the points of the line $\ell$ do not belong to the set, and the other boundary points do belong).

Note that due to symmetry, the segment of the circle located above the chord $A B$ is equal to the segment of the circle located above the chord $A O$. Therefore, the area of the figure $M$ is equal to the area of the triangle $A C O$, i.e., $\frac{1}{2} \cdot 4 \cdot 4=8$.