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

$$
\left\{\begin{array}{l}
x+y+|x-y| \leqslant 0 \\
\frac{x^{2}+6 x+y^{2}-8 y}{x+3 y+6} \geqslant 0
\end{array}\right.
$$

Sketch the figure $M$ and find its area.
Answer: 3.

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

Consider the second inequality. It can be written as $\frac{(x+3)^{2}+(y-4)^{2}-25}{x+3 y+6} \geqslant 0$. The numerator of the fraction on the left side of the inequality is zero on the circle of radius 5 centered at the point $Q(-3 ; 4)$ (let's call it $\omega$). The denominator of the fraction is zero on the line $y=-2-\frac{x}{3}$ (let's call it $\ell$). The points of intersection of the circle and the line are determined from the system of equations

$$
\left\{\begin{array} { l } 
{ x = - 3 y - 6 , } \\
{ x ^ { 2 } + 6 x + y ^ { 2 } - 8 y = 0 }
\end{array} \Leftrightarrow \left\{\begin{array} { l } 
{ x = - 3 y - 6 } \\
{ ( 3 y + 6 ) ^ { 2 } - 6 ( 3 y + 6 ) + y ^ { 2 } - 8 y = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=-3 y-6 \\
y^{2}+y=0
\end{array}\right.\right.\right.
$$

from which we obtain two points $A(-3 ;-1)$ and $B(-6 ; 0)$. 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 $(0 ;-2)$).

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 below the line $\ell$ (the numerator and denominator are negative);
- outside the circle $\omega$ at points located above the line $\ell$ (the numerator and denominator are positive).

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 below 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 below the chord $A B$ is equal to the segment of the circle located below the chord $A O$. Therefore, the area of the figure $M$ is equal to the area of the triangle $A C O$, i.e., $\frac{1}{2} \cdot 3 \cdot 2=3$.