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

$$
\left\{\begin{array}{l}
x \cos a - y \sin a - 3 \leqslant 0 \\
x^{2} + y^{2} - 8 x + 2 y - b^{2} - 6 b + 8 = 0
\end{array}\right.
$$

has at least one solution for any value of the parameter $a$.
Answer. $b \in(-\infty ;-\sqrt{17}] \cup[\sqrt{17}-6 ;+\infty)$.

Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a-y \sin a-3=0$ is 3, and the point $(0 ; 0)$ satisfies this inequality. Therefore, the inequality defines a half-plane containing the point $(0 ; 0)$, with the boundary being a line tangent to the circle $x^{2}+y^{2}=9$.

The equation of the given system can be transformed into the form $(x-4)^{2}+(y+1)^{2}=(b+3)^{2}$. It represents a circle $\Omega(b)$ with center $(4 ;-1)$ and radius $|b+3|$ (or the point $(4 ;-1)$ when $b=-3$).

For the system to have a solution for any value of the parameter $a$, it is required that the circle $\Omega(b)$ intersects any of the half-planes defined by the inequality of the system. Let $r_{0}$ be the radius of the circle $\Omega(b)$ that is externally tangent to the circle $x^{2}+y^{2}=9$. Then, the condition is satisfied by all radii in the interval $\left[r_{0} ;+\infty\right)$.

For circles that are externally tangent, the sum of the radii is equal to the distance between the centers. From this, we get that $r_{0}=\sqrt{17}-3$, so $|b+3| \geqslant \sqrt{17}-3 \Leftrightarrow b \in(-\infty ;-\sqrt{17}] \cup[\sqrt{17}-6 ;+\infty)$.