7. [6 points] Find all values of the parameter $a$, for each of which the system

$$
\left\{\begin{array}{l}
|y+x+8|+|y-x+8|=16 \\
(|x|-15)^{2}+(|y|-8)^{2}=a
\end{array}\right.
$$

has exactly two solutions.
Answer: $a \in\{49 ; 289\}$.

Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To open the absolute values, we need to find the sets of points where the expressions under the absolute values are zero. These are the lines $y+x+8=0$ and $y-x+8=0$. They divide the plane into 4 parts, and in each of these parts, the signs of the expressions under the absolute values are constant. To determine them, we can choose a point in each of the four parts and find the signs of the expressions at these points.

![](https://cdn.mathpix.com/cropped/2024_05_06_c4692c563a78e68b8782g-09.jpg?height=1085&width=1108&top_left_y=243&top_left_x=491)

Fig. 4: variant 4, problem 7

Consider the region located above both lines. For example, the point $(0 ; 10)$ lies in this region. By substitution, it is easy to verify that both expressions $y+x+8$ and $y-x+8$ are positive at this point. Thus, the equation takes the form $(y+x+8)+(y-x+8)=16$, from which $y=0$. Taking into account the considered constraints, the segment with endpoints at points $A(8 ; 0)$ and $D(-8 ; 0)$ fits. Similarly, we consider the other three cases, and in the end, we obtain the boundaries of the square $K$ with vertices at points $A(8 ; 0), B(8 ;-16), C(-8 ;-16)$, and $D(-8 ; 0)$. This figure does not intersect the half-plane $y>0$, so we can assume that $y \leqslant 0$. Taking this remark into account, the second equation can be written as $(|x|-15)^{2}+(y+8)^{2}=a$ (by opening the absolute value for the variable $y$). Denote the set of points defined by this equation as $\Phi(a)$. If $a0$.

For $x \geqslant 0$, the equation takes the form $(x-15)^{2}+(y+8)^{2}=a$, and we obtain a circle of radius $\sqrt{a}$ centered at the point $(15 ;-8)$ (or its part lying in the half-plane $x \geqslant 0$, if the entire circle does not fit in this half-plane). Since the equation is invariant under the substitution of $x$ with $(-x)$, the set $\Phi(a)$ is symmetric with respect to the $O y$ axis. Thus, $\Phi(a)$ is the union of the above circle (or its part) and the circle obtained by reflecting the already constructed circle with respect to the $O y$ axis.

If $0289$, the arc of the circle $(x-15)^{2}+(y+8)^{2}=a, x \geqslant 0$ does not intersect the sides of the square $K$, and the system of equations has no solutions. Therefore, the system of equations has exactly two solutions only for $a=49$ and $a=289$.

The problem is considered fully solved (and the maximum number of points is awarded) only if all necessary transformations are provided in the solution text and all logical steps are fully explained; the obtained answers are also simplified.

The presence of a correct answer does not guarantee a positive score for the problem. A correct answer without justification does not add points.

A full score (indicated in parentheses after the problem number) is given for a correct and justified solution. Some partial progress is evaluated according to the instructions. In other cases, the score is at the discretion of the checker.

One point is deducted for an arithmetic error that does not significantly affect the solution process.