5. For what values of $a$ does the system $\left\{\begin{array}{c}(x-7 \cos a)^{2}+(y-7 \sin a)^{2}=1 \\ |x|+|y|=8\end{array}\right.$ have a unique solution?
Solution. The set of points on the plane, with coordinates $(x ; y)$ that satisfy the first equation of the system, lie on a circle $K_{a}$ of radius 1 centered at the point $O(7 \cos a ; 7 \sin a)$. As $a$ changes, the center moves along a circle of radius 7 centered at the origin.

![](https://cdn.mathpix.com/cropped/2024_05_06_6cabcf1712d23d5ceb2bg-08.jpg?height=380&width=394&top_left_y=1398&top_left_x=999)

The set of points on the plane, with coordinates $(x ; y)$ that satisfy the second equation of the system, form the boundary of the square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6cabcf1712d23d5ceb2bg-08.jpg?height=554&width=531&top_left_y=1916&top_left_x=817)

When $a=0$, point $B$ lies on the circle $K_{0}$, and the system has three solutions. For $a \in\left(0 ; a_{1}\right)$, where the value $a=a_{1}$ corresponds to the circle $K_{a}$ being tangent to side $B C$ of the square, vertex $B$ is outside the unit circle, and the system has 4 solutions. When $a=a_{1}$, the system has 3 solutions. On the interval $a \in\left(a_{1} ; a_{2}\right)$, where the value $a=a_{2}$ corresponds to the circle $K_{a}$ being tangent to side $A B$ of the square, the system has two solutions, corresponding to two points of intersection of the circle with side $A B$. When $a=a_{2}$, the system has a unique solution. On the interval $a \in\left(a_{2} ; \frac{\pi}{2}-a_{2}\right)$, the circle $K_{a}$ is outside the square, and the system has no solutions. When $a_{3}=\frac{\pi}{2}-a_{2}$, the system has a unique solution. Then, by the symmetry of the phase portrait relative to the bisector of the first quadrant, on the interval $\left(a_{3} ; \frac{\pi}{2}-a_{1}\right)$, the system will have two solutions, and at $a_{4}=\frac{\pi}{2}-a_{1}$, it will have three solutions. Finally, for $a \in\left(a_{4} ; \frac{\pi}{2}\right)$, the system has 4 solutions, and for $a=\frac{\pi}{2}$, it has three solutions. It can be noted that $T=\frac{\pi}{2}$ is the period for the solution of the problem. Thus, the solution to the problem variant 1 is the series $a=a_{2}+\frac{\pi k}{2}, k \in Z$ and $a=\frac{\pi}{2}-a_{2}+\frac{\pi k}{2}, k \in Z$, which combine into the series $a= \pm a_{2}+\frac{\pi k}{2}, k \in Z$.

Calculation of the value of $a_{2}$:

![](https://cdn.mathpix.com/cropped/2024_05_06_6cabcf1712d23d5ceb2bg-09.jpg?height=391&width=411&top_left_y=1341&top_left_x=914)

$N$ - the point of tangency of the circle with the line $A B$, $M(x ; y)$ - the point of intersection of the line $Q O$ with the line $A B$.

$x=Q M \cos a, y=Q M \sin a \rightarrow x+y=8=Q M(\sin a+\cos a) \rightarrow Q M=\frac{4 \sqrt{2}}{\sin \left(\frac{\pi}{4}+a\right)}$

$a=a_{2}=\arcsin \frac{4 \sqrt{2}+1}{7}-\frac{\pi}{4}$. Considering the symmetry of the square, the system has a unique solution at $a= \pm\left(\arcsin \frac{4 \sqrt{2}+1}{7}-\frac{\pi}{4}\right)+\frac{\pi k}{2}= \pm \arcsin \frac{4 \sqrt{2}+1}{7}+\frac{(2 k-1) \pi}{4}, k \in Z$.

Answer: $a= \pm \arcsin \frac{4 \sqrt{2}+1}{7}+\frac{(2 k-1) \pi}{4}, k \in Z$.