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

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) $S=5 \pi-5 \operatorname{arctg} 2+2$, b) $a=-10, a=2$.

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

$$
\left\{\begin{array} { l } 
{ 4 - 2 x - x ^ { 2 } - y ^ { 2 } \geq 0 , } \\
{ - 2 x \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+1)^{2}+y^{2} \leq 5 \\
x \leq 0
\end{array}\right.\right.
$$

The first inequality defines a circle of radius $\sqrt{5}$ centered at $(-1 ; 0)$, and the entire system defines the part of this circle lying in the half-plane $x \leq 0$. 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 $x \geq 0$. Since the central angle of this segment is $2 \operatorname{arctg} 2$, we get that its area $S$ is $5 \pi-5 \operatorname{arctg} 2+2$.

b) Consider the second equation of the original system. Rewrite it as $a(3 x-1)+15 x-3 y-4=0$. If we substitute $x=\frac{1}{3}$ into it, we get that $y=\frac{1}{3}$. Thus, this equation defines a line passing through the point $\left(\frac{1}{3} ; \frac{1}{3}\right)$. This point lies in the segment cut off from the circle.

“Phystech-2016", mathematics, solutions to ticket 24

This means that the system has exactly one solution when the line passes through one of the "vertices" of the segment - the point $M(0 ; 2)$ or the point $P(0 ;-2)$. Substituting the coordinates of the points into the equation of the line, we get:

$$
\begin{aligned}
& M(0 ; 2) \rightarrow 0=6+a+4 \Leftrightarrow a=-10 \\
& P(0 ;-2) \rightarrow 0=-6+a+4 \Leftrightarrow a=2
\end{aligned}
$$