5. [6 points] Find all values of the parameter $a$ for which the system

$$
\left\{\begin{array}{l}
x=|y-\sqrt{a}|+\sqrt{a}-4 \\
(|x|-6)^{2}+(|y|-8)^{2}=100
\end{array}\right.
$$

has exactly three solutions.

![](https://cdn.mathpix.com/cropped/2024_05_06_ffd4340bf000b94108d2g-33.jpg?height=1093&width=1088&top_left_y=1184&top_left_x=495)

Fig. 5: variant 14, problem 5
Answer: $a \in\{4 ; 64 ; 51+10 \sqrt{2}\}$.

Consider the second equation of the system. It is invariant under the replacement 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 expanding the absolute values, we get $(x-6)^{2}+(y-8)^{2}=100$. This equation defines a circle with center $(6 ; 8)$ 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 (see figure).

The geometric locus of points defined by the first equation represents the union of two rays $\ell_{1}$ and $\ell_{2}$ with the origin at the point ($\sqrt{a}-4, \sqrt{a}$), forming angles of $-\frac{\pi}{4}$ and $\frac{\pi}{4}$ with the positive direction of the $O x$ axis, respectively.

Note that the ray $\ell_{2}$ is part of the line $y=x+4$ for any $a$ and does not intersect the half-plane $x(5 \sqrt{2}+1)^{2}$. The ray $\ell_{1}$ can intersect only the arc of the circle lying in the first quadrant, and the total number of intersection points of the graphs does not exceed two. Thus, $a \in\left\{4 ; 64 ;(5 \sqrt{2}+1)^{2}\right\}$.

The value $a=(5 \sqrt{2}+1)^{2}$, corresponding to the tangency of $\ell_{1}$ and $\omega$, can be found, for example, as follows. Let $Q(6 ;-8)$ be the center of the circle containing the arc $\omega, P$ be the point of tangency of $\ell_{1}$ and $\omega$. Since the slope of $\ell_{1}$ is $(-1)$, the slope of the radius $Q P$ is 1, which means that the coordinates of point $P$ are $\left(6+10 \cos \frac{\pi}{4} ;-8+10 \sin \frac{\pi}{4}\right)$. Since the ray $\ell_{1}$ with the equation $x=$ $-(y-\sqrt{a})+\sqrt{a}-4$ passes through point $P$, we get $6+5 \sqrt{2}=2 \sqrt{a}+4-5 \sqrt{2}$, from which $a=(5 \sqrt{2}+1)^{2}$.