Problem 898

Answer. $b \in(-\infty ;-8-\sqrt{26}] \cup[\sqrt{26} ;+\infty)$.

Solution. Consider the inequality of the given system. For any value of the parameter $a$, the distance from the origin to the line $x \cos a+y \sin a+4=0$ 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 $x^{2}+y^{2}=16$.

The equation of the given system can be transformed into the form $(x+5)^{2}+(y+1)^{2}=(b+4)^{2}$. It defines a circle $\Omega(b)$ with center $(-5 ;-1)$ and radius $|b+4|$ (or the point $(-5 ;-1)$ when $b=-4$).

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 $x^{2}+y^{2}=16$ internally (i.e., the circle $x^{2}+y^{2}=16$ is inside the circle $\Omega(b)$). Then, the condition is satisfied by all radius values from the interval $\left[r_{0} ;+\infty\right)$.

For circles touching internally, the difference in radii equals the distance between the centers. From this, we get that $r_{0}=\sqrt{26}+4$, so $|b+4| \geqslant \sqrt{26}+4 \Leftrightarrow b \in(-\infty ;-8-\sqrt{26}] \cup[\sqrt{26} ;+\infty)$.