6. Find all values of the parameter $a$, for each of which the system of equations

$$
\left\{\begin{array}{l}
(|y+9|+|x+2|-2)\left(x^{2}+y^{2}-3\right)=0 \\
(x+2)^{2}+(y+4)^{2}=a
\end{array}\right.
$$

has exactly three solutions.
Answer: $a=9, a=23+4 \sqrt{15}$.

Solution. The first equation of the given system is equivalent to the combination of two equations $|y+9|+|x+2|=2$ and $x^{2}+y^{2}=3$. The first of these defines a square $G$ with center $(-2, -9)$, the diagonals of which are 4 and parallel to the coordinate axes. The second defines a circle $S$ with center $(0, 0)$ and radius $\sqrt{3}$.

The second equation of the original system, for $a>0$, defines a circle $\Omega$ with center $(-2, -4)$ and radius $R=\sqrt{a}$ (for $a<\sqrt{3}$, the circle $\Omega$ is completely inside the circle $S$ and the system has no solutions; for $\sqrt{3}<a<20$, the circle $\Omega$ intersects the circle $S$ at two points, and the system has 2 solutions; for $a=20$, the circle $\Omega$ is tangent to the circle $S$ at one point, and the system has 1 solution; for $a>20$, the circle $\Omega$ intersects the square $G$ at one point, and the system has 1 solution).

Thus, $R=3$ and $R=\sqrt{20}+\sqrt{3}$ are suitable. Then $a=9$ and $a=23+4 \sqrt{15}$.