6. (7 points) A radar station (Radar) during a circular scan detected an unidentified flying object (UFO) hovering motionless on the horizon, defined on the map by the set $25 x^{2}+x y+y^{2}+16 x+2 y+3 \leq 0$. Find the extreme azimuths of the UFO, i.e., within what angle with the vertex at the origin (the location of the Radar) is this object, which gives a glow on the screens of the circular scan. Estimate $\min R$ - the minimum distance of the UFO from the Radar. Can $R$ be less than 0.5 conditional units of length?
## Solution.

The UFO, defined on the map by the inequality $25 x^{2}+x y+y^{2}+16 x+2 y+3 \leq 0$, (1) is located in one of the quadrants, as it does not intersect the coordinate axes.

For $x=0$, from (1) we get $y^{2}+2 y+3>0$ for any $y$.

For $y=0$, from (1) we get $25 x^{2}+16 x+3>0, x \in \boldsymbol{R}$.

The shortest azimuths are found from the system

$$
\left\{\begin{array}{c}
25 x^{2}+x y+y^{2}+16 x+2 y+3=0 \\
y=k x
\end{array}\right.
$$

when it has a unique solution:

$\left(k^{2}+k+25\right) x^{2}+2(k+8) x+3=0$ (3)

$D=0 \Leftrightarrow 2 k^{2}-13 k+11=0 \Leftrightarrow k_{1,2}=\frac{13 \pm 9}{4}=1 \text{ or } \frac{11}{2}$

Then the UFO, obviously, cannot be in the second and fourth quadrants, since for $x \geq 0, y \geq 0$ the inequality (1) is not satisfied.

Therefore, the UFO is located in the third quadrant between the angles (extreme azimuths) $\alpha_{1}=\pi+\operatorname{arctg} \frac{11}{2}$
$\alpha_{2}=\pi+\frac{\pi}{4}$

Therefore, from the origin, the UFO is visible at an angle

$\alpha=\alpha_{1}-\alpha_{2}=\operatorname{arctg} \frac{11}{2}-\operatorname{arctg} \frac{\pi}{4}$

For $k=1$, equation (3) is equivalent to $27 x^{2}+18 x+3=0, x=-\frac{1}{3}$.

Then $y=k x=-\frac{1}{3}$.

Therefore, $\operatorname{minR} \leq \sqrt{\left(-\frac{1}{3}\right)^{2}+\left(-\frac{1}{3}\right)^{2}}=\frac{\sqrt{2}}{3} \approx 0.47<0.5$

Other estimates of $\min R$ (for example, at $k=2$ or $k=3$) will be worse.