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

$$
\left\{\begin{array}{l}
x \cos a + y \sin a + 3 \leqslant 0 \\
x^{2} + y^{2} + 8x - 4y - b^{2} + 6b + 11 = 0
\end{array}\right.
$$

has at least one solution for any value of the parameter $a$.
Answer. $b \in(-\infty ;-2 \sqrt{5}] \cup[6+2 \sqrt{5} ;+\infty)$.

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 $(0 ; 0)$ does not satisfy this inequality. Therefore, the inequality defines a half-plane that does not contain 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 $(x+4)^{2}+(y-2)^{2}=(b-3)^{2}$. It represents a circle $\Omega(b)$ with center $(-4 ; 2)$ and radius $|b-3|$ (or the point $(-4 ; 2)$ 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 touches the circle $x^{2}+y^{2}=9$ internally (i.e., the circle $x^{2}+y^{2}=9$ is inside the circle $\Omega(b)$). Then, the condition is satisfied by all radii in the interval $\left[r_{0} ;+\infty\right)$.

For circles that touch each other internally, the difference in radii equals the distance between the centers. From this, we get that $r_{0}=\sqrt{20}+3$, so $|b-3| \geqslant \sqrt{20}+3 \Leftrightarrow b \in(-\infty ;-2 \sqrt{5}] \cup[6+2 \sqrt{5} ;+\infty)$.