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

$$
\left\{\begin{array}{l}
(|y+2|+|x-11|-3)\left(x^{2}+y^{2}-13\right)=0 \\
(x-5)^{2}+(y+2)^{2}=a
\end{array}\right.
$$

has exactly three solutions.
Answer: $a=9, a=42+2 \sqrt{377}$.

Solution. The first equation of the given system is equivalent to the combination of two equations $|y+2|+|x-11|=3$ and $x^{2}+y^{2}=13$. The first of these equations defines a square $G$ with center at $(11, -2)$, the diagonals of which are 6 and parallel to the coordinate axes. The second equation defines a circle $S$ with center at $(0, 0)$ and radius $\sqrt{13}$.

The second equation of the original system, for $a>0$, defines a circle $\Omega$ with center at $(5, -2)$ and radius $R=\sqrt{a}$ (for $a\sqrt{29}+\sqrt{13}$), i.e., the system has 1 solution.

Thus, $R=3$ and $R=\sqrt{29}+\sqrt{13}$ are suitable. Therefore, $a=9$ and $a=42+2 \sqrt{377}$.