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

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

has at least one solution for any value of the parameter $a$.
Answer. $b \in(-\infty ; 4-\sqrt{10}] \cup[\sqrt{10} ;+\infty)$.

Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a-2=0$ is 2, 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 $x^{2}+y^{2}=4$.

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

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 $x^{2}+y^{2}=4$. 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 $r_{0}=\sqrt{10}-2$, so $|b-2| \geqslant \sqrt{10}-2 \Leftrightarrow b \in(-\infty ; 4-\sqrt{10}] \cup[\sqrt{10} ;+\infty)$.