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

$$
\left\{\begin{array}{l}
|y-6-x|+|y-6+x|=12 \\
(|x|-8)^{2}+(|y|-6)^{2}=a
\end{array}\right.
$$

has exactly two solutions.
Answer: $a \in\{4 ; 100\}$.

Solution. Consider the first equation of the system and represent the set of its solutions on the coordinate plane. To remove the absolute values, we find the sets of points where the expressions under the absolute values are zero. These are the lines $y-6-x=0$ and $y-6+x=0$. They divide the plane

![](https://cdn.mathpix.com/cropped/2024_05_06_c4692c563a78e68b8782g-04.jpg?height=1082&width=1102&top_left_y=244&top_left_x=480)

Fig. 2: variant 3, problem 7

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. Consider the region located below both lines. For example, the point $(0 ;-10)$ lies in this region. By substitution, it is easy to verify that both expressions $y-6-x$ and $y-6+x$ are negative at this point. Thus, the equation takes the form $-(y-6-x)-(y-6+x)=12$, from which $y=0$. Taking into account the considered constraints, the segment with endpoints at points $A(6 ; 0)$ and $D(-6 ; 0)$ fits. Similarly, we consider the other three cases, and in the end, we obtain the boundaries of the square $K$ with vertices at points $A(6 ; 0), B(6 ; 12), C(-6 ; 12)$, and $D(-6 ; 0)$. This figure does not intersect the half-plane $y>0$.

For $x \geqslant 0$, the equation takes the form $(x-8)^{2}+(y-6)^{2}=a$, and we obtain a circle of radius $\sqrt{a}$ centered at the point $(8 ; 6)$ (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 one with respect to the $O y$ axis.

If $0<a<4$, the arc of the circle $(x-8)^{2}+(y-6)^{2}=a, x \geqslant 0$ does not intersect the sides of the square $K$, and the system of equations has no solutions. If $a=4$, the arc of the circle intersects the sides of the square $K$ at exactly two points. If $4<a<100$, the arc of the circle intersects the sides of the square $K$ at more than two points. If $a=100$, the arc of the circle intersects the sides of the square $K$ at exactly two points. If $a>100$, the arc of the circle $(x-8)^{2}+(y-6)^{2}=a, x \geqslant 0$ does not intersect the sides of the square $K$, and the system of equations has no solutions. Thus, the system of equations has exactly two solutions only for $a=4$ and $a=100$.

## VARIANT 4