6. Given the system of equations $\left\{\begin{array}{l}\left\lvert\, \begin{array}{l}16+6 x-x^{2}-y^{2}|+| 6 x \mid=16+12 x-x^{2}-y^{2} \\ (a+15) y+15 x-a=0 .\end{array}\right.\end{array}\right.$

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) $25 \pi-25 \arcsin 0.8+12$, b) $a=-20, a=-12$.

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 side). Therefore, the first equation is equivalent to the system of inequalities

$$
\left\{\begin{array}{l}
16 + 6 x - x^2 - y^2 \geq 0 \\
6 x \geq 0
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-3)^2 + y^2 \leq 25 \\
x \geq 0
\end{array}\right.\right.
$$

The first inequality defines a circle of radius 5 centered at $(3 ; 0)$, and the entire system defines the part of this circle lying in the half-plane $x \geq 0$. 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 $x \leq 0$. Since the central angle of this segment is $2 \arcsin 0.8$, we get that its area $S$ is $25 \pi-25 \arcsin 0.8+12$.

b) Consider the second equation of the original system. Rewrite it as $a(y-1)+15 x+15 y=0$. If we substitute $y=1$ into it, we get that $x=-1$. Thus, this equation defines a line passing through the point $(-1 ; 1)$. 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 - the point $M(0 ; 4)$ or the point $P(0 ;-4)$. Substituting the coordinates of the points into the equation of the line, we get:

“Fiztekh-2016", mathematics, solutions of ticket 21

$$
\begin{aligned}
M(0 ; 4) & \rightarrow 4(a+15)-a=0 \Leftrightarrow a=-20 \\
P(0 ;-4) & \rightarrow-4(a+15)-a=0 \Leftrightarrow a=-12
\end{aligned}
$$