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

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

has at least one solution $(x ; y)$.
Answer. $b \leq 2 \sqrt{2}+\frac{1}{4}$.

Solution. The second equation of the system can be transformed into the form $(x-2 a)^{2}+(y-2 a)^{2}=2^{2}$, hence it represents a circle of radius 2 with center at $(2 a ; 2 a)$. For all possible $a \in \mathrm{R}$, the graphs of these functions sweep out the strip $x-2 \sqrt{2} \leq y \leq x+2 \sqrt{2}$.

For the condition of the problem to be satisfied, it is necessary and sufficient that the parabola defined by the first equation has at least one common point with this strip.

Let's find the value of the parameter $b$ for which the parabola touches the lower boundary of the strip, i.e., the line $y=x-2 \sqrt{2}$. This means that the equation $-b-x^{2}=x-2 \sqrt{2}$ has exactly one solution, from which we get $b=2 \sqrt{2}+\frac{1}{4}$. In this case, the ordinate of the vertex of the parabola is $y_{0}=-2 \sqrt{2}-\frac{1}{4}$. All values of $b$ for which $y_{\mathrm{B}} \geq y_{0}$ are suitable, i.e., $-b \geq-2 \sqrt{2}-\frac{1}{4}, b \leq 2 \sqrt{2}+\frac{1}{4}$