6. a) On the coordinate plane, plot the figure $\Phi$, the coordinates of the points of which satisfy the system of inequalities

$$
\left\{\begin{array}{l}
y^{2}-x^{2} \leqslant 3(x+y) \\
x^{2}+y^{2} \leqslant 6 y-6 x-9
\end{array}\right.
$$

b) Find the area of the figure $\Phi$ and the distance from the point $T(-6 ; 0)$ to the nearest point of the figure $\Phi$.
Answer. $S=\frac{9 \pi}{2} ; \rho=3 \sqrt{2}-3$.

Solution. Transform the first inequality: $(y-x)(y+x)-3(y+x) \leqslant 0 \Leftrightarrow(y+x)(y-x-3) \leqslant 0$. It defines two vertical angles, the boundaries of which are the lines $y=x$ and $y=2-x$ (the points $( \pm 10 ; 0)$ lie inside these angles). The second inequality can be rewritten as $(x+3)^{2}+(y-3)^{2} \leqslant 9$. It defines a circle with center $G(-3 ; 3)$ and radius 3. The intersection of these two sets is the desired set.

The figure $\Phi$ constitutes exactly half of a circle with radius 3, so its area is $\frac{1}{2} \cdot 9 \pi=\frac{9 \pi}{2}$. Mark the point $E$ of intersection of the segment $G T$ with the circle. $T E$ is the shortest distance from point $T$ to the figure $\Phi . T E=T G-G E=3 \sqrt{2}-3$.