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

$$
\left\{\begin{array}{l}
x^{2}+(y-a)^{2}=64 \\
(|x|-6)^{2}+(|y|-8)^{2}=100
\end{array}\right.
$$

has exactly two solutions.
Answer: $a \in\{-8-12 \sqrt{2}\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{12 \sqrt{2}+8\}$.

Solution. Consider the second equation of the system. It is invariant under the substitution of $x$ with $(-x)$ and/or $y$ with $(-y)$. This means that the set of points defined by this equation is symmetric with respect to both coordinate axes. In the first quadrant (including its boundaries), by removing the absolute values, we get $(x-6)^{2}+(y-8)^{2}=100$. This equation defines a circle with center $A(6 ; 8)$ and radius 10. In the first quadrant, there is an arc of this circle and the point $O(0 ; 0)$. By reflecting this arc symmetrically with respect to the origin and both coordinate axes, we obtain the set of points defined by the second equation (denote this set as $M$ - see the figure).

The geometric locus of points defined by the first equation is a circle $\Omega$ with radius 8 and center at point $F(0 ; a)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c4692c563a78e68b8782g-15.jpg?height=1079&width=1088&top_left_y=246&top_left_x=495)

Fig. 2: variant 11, problem 7

Consider positive values of $a$.

If $0<a<8$, the circle $\Omega$ has no common points with the set $M$.

If $a=8$, we get two solutions (points $(0 ; 0)$ and $(0 ; 16)$).

If $8<a<24$, the system has two solutions.

When $a=24$, the circle $\Omega$ has three common points with the set $M$ (depicted with a dashed line in the figure).

If $24<a<8+12 \sqrt{2}$, $\Omega$ intersects the figure $M$ twice in the first quadrant and twice in the fourth quadrant - a total of four solutions.

Finally, when $a=8+12 \sqrt{2}$, $\Omega$ and $M$ have exactly two common points (the case of tangency - see the drawing). The value $a=8+12 \sqrt{2}$, at which the circles are tangent, can be found as follows. Let $P(0 ; a)$ be the position of the center of the circle at which the tangency occurs, and $H(8 ; 0)$ be the foot of the perpendicular dropped from point $A$ to the y-axis. Then $A H=6$; the segment $A P$ is the distance between the centers of the tangent circles, so it is equal to the sum of the radii, i.e., $A P=10+8=25$. Then, by the Pythagorean theorem in triangle $A H P$, we get $H P=$ $\sqrt{18^{2}-6^{2}}=288$. Therefore, the point $P$ has an ordinate $8+\sqrt{288}=8+12 \sqrt{2}, a=8+12 \sqrt{2}$.

Thus, among the positive $a$, the suitable values are $a \in[8 ; 24) \cup\{12 \sqrt{2}\}$. By symmetry, for negative values of the parameter, we get $a \in\{-12 \sqrt{2}\} \cup(-24 ;-8]$. It is also clear that the value $a=0$ does not work. Therefore, $a \in\{-12 \sqrt{2}\} \cup(-24 ;-8] \cup[8 ; 24) \cup\{12 \sqrt{2}\}$.

## VARIANT 12