6. Find all values of the parameter $a$ such that the system

$$
\left\{\begin{array}{l}
\arccos \left(\frac{4+y}{4}\right)=\arccos (x-a) \\
x^{2}+y^{2}-4 x+8 y=b
\end{array}\right.
$$

has no more than one solution for any value of the parameter $b$.
Answer. $a \in(-\infty ;-15] \cup[19 ;+\infty)$.

Solution. The first equation on the domain of definition is equivalent to the equation $\frac{4+y}{4}=x-a, y=4 x-4 a-4$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4+y}{4} \leqslant 1,-8 \leqslant y \leqslant 0$. Thus, the first equation defines a segment $A B$ on the plane, the position of which depends on the parameter $a$.

The second equation can be rewritten as $(x-2)^{2}+(y+4)^{2}=b+20$ - this is the equation of a circle with center $M(2 ;-4)$ and radius $\sqrt{b+20}$ (it can also be a point or an empty set, but then for any $a$ there is no more than one solution).

The system has no more than one solution for any $b$ if and only if the perpendicular dropped from $M$ to the line containing the segment $A B$ does not fall into an internal point of the segment (if the circle intersects the line, then the points of intersection are on opposite sides of the projection of the center of the circle onto the line).

Let's write the equation of the line passing through $M$ and perpendicular to $A B$. Since the product of the slopes of mutually perpendicular lines is -1, its slope is $-\frac{1}{4}$ and its equation is $y+4=-\frac{1}{4}(x-2)$, i.e., $y=-\frac{x}{4}-\frac{7}{2}$. The ordinate of the point of intersection of this line and the line $A B$ can be found from the system of equations $y=-\frac{x}{4}-\frac{7}{2}, y=4 x-4 a-4$ - this is $y_{0}=-\frac{4 a+60}{17}$. For this point not to be an internal point of the segment, it is necessary and sufficient that $y_{0} \notin(-8 ; 0)$, from which $a \leqslant-15$ or $a \geqslant 19$.