6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}9-x^{2}-y^{2}-2 y|+|-2 y \mid=9-x^{2}-y^{2}-4 y \\ 15 y+3 a=(4 a+15) x .\end{array}\right.\end{array}\right.$

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) $S=10 \pi+3-10 \operatorname{arctg} 3$, b) $a=-5, a=-3$.

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

$$
\left\{\begin{array} { l } 
{ 9 - x ^ { 2 } - y ^ { 2 } - 2 y \geq 0 , } \\
{ - 2 y \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(y+1)^{2}+x^{2} \leq 10 \\
y \leq 0
\end{array}\right.\right.
$$

The first inequality defines a circle of radius $\sqrt{10}$ centered at $(0 ;-1)$, and the entire system defines the part of this circle lying in the half-plane $y \leq 0$. The area of this segment is the difference between the area of the circle and the area of the segment of this circle lying in the half-plane $y \geq 0$. Since the central angle of this segment is $2 \operatorname{arctg} 3$, we get that its area $S$ is $10 \pi+3-10 \operatorname{arctg} 3$.

b) Consider the second equation of the original system. Rewrite it as $a(4 x-3)+15 x-15 y=0$. If we substitute $x=\frac{3}{4}$ into it, we get that $y=\frac{3}{4}$. Thus, this equation defines a line passing through the point $\left(\frac{3}{4} ; \frac{3}{4}\right)$. This point lies in the segment cut off from the circle.

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

$$
\begin{aligned}
M(3 ; 0) & \rightarrow 3 a=3(4 a+15) \Leftrightarrow a=-5 \\
P(-3 ; 0) & \rightarrow 3 a=-3(4 a+15) \Leftrightarrow a=-3
\end{aligned}
$$