6. Find all values of the parameter $a$ for which there exists a value of the parameter $b$ such that the system

$$
\left\{\begin{array}{l}
\arcsin \left(\frac{a-y}{3}\right)=\arcsin \left(\frac{4-x}{4}\right) \\
x^{2}+y^{2}-8 x-8 y=b
\end{array}\right.
$$

has exactly two solutions.
Answer. $a \in\left(-\frac{13}{3} ; \frac{37}{3}\right)$.

Solution. The first equation on the domain of definition is equivalent to the equation $\frac{a-y}{3}=\frac{4-x}{4}, y=\frac{3 x}{4}-3+a$. The domain of definition is determined by the inequality $-1 \leqslant \frac{4-x}{4} \leqslant 1,0 \leqslant x \leqslant 8$. 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 $(x-4)^{2}+(y-4)^{2}=b+32$ - this is the equation of a circle with center $M(4 ; 4)$ and radius $\sqrt{b+32}$ (it can also be a point or an empty set, but these cases do not interest us, as then the system has fewer than two solutions).

The system can have two solutions for some $b$ if and only if the perpendicular dropped from $M$ to the line containing the segment $A B$ falls on 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 $-\frac{4}{3}$ and its equation is $y-4=-\frac{4}{3}(x-4)$, i.e., $y=-\frac{4}{3} x+\frac{28}{3}$. The abscissa of the point of intersection of this line and the line $A B$ can be found from the system of equations $y=-\frac{4}{3} x+\frac{28}{3}, y=\frac{3 x}{4}-3+a-$ this is $x_{0}=\frac{148-12 a}{25}$.

For this point to be an internal point of the segment, it is necessary and sufficient that $0<x_{0}<8$, from which $-\frac{13}{3}<a<\frac{37}{3}$.