6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}9+8 y-x^{2}-y^{2}|+| 8 y \mid=16 y+9-x^{2}-y^{2} \\ (a+4) x-13 y+a=0 .\end{array}\right.\end{array}\right.$

a) Plot on the plane $(x ; y)$ the set of points satisfying the first equation of the system, and find the area of the resulting figure.

b) Find all values of the parameter $a$ for which the system has exactly one solution.
Answer: a) $25 \pi-25 \arcsin 0.6+12$, b) $a=-6, a=-3$.

Solution. a) Note that the equality $|a|+|b|=a+b$ holds if and only if the numbers $a$ and $b$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities

$$
\left\{\begin{array}{l}
9 + 8 y - x ^ { 2 } - y ^ { 2 } \geq 0 , \\
8 y \geq 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(y-4)^{2}+x^{2} \leq 25, \\
y \geq 0 .
\end{array}\right.\right.
$$

The first inequality defines a circle of radius 5 centered at $(0 ; 4)$, and the entire system defines the part of this circle lying in the half-plane $y \geq 0$. The area of this segment is equal to the difference between the area of the circle and the area of the segment of this circle lying in the half-plane $y \leq 0$. Since the central angle of this segment is $2 \arcsin 0.6$, we get that its area $S$ is $25 \pi-25 \arcsin 0.6+12$.

b) Consider the second equation of the original system. Rewrite it as $a(x+1)+4 x-13 y=0$. If we substitute $x=-1$ into it, we get that $y=-\frac{4}{13}$. Thus, this equation defines a line passing through the point $\left(-1 ;-\frac{4}{13}\right)$. This point lies in the segment cut off from the circle.

Therefore, the system has exactly one solution when the line passes through one of the "vertices" of the segment - point $M(3 ; 0)$ or point $P(-3 ; 0)$. Substituting the coordinates of the points into the equation of the line, we get:

$$
\begin{gathered}
M(3 ; 0) \rightarrow 3(a+4)+a=0 \Leftrightarrow a=-3 \\
P(-3 ; 0) \rightarrow-3(a+4)+a=0 \Leftrightarrow a=-6
\end{gathered}
$$