4. For what values of the parameter $a$ does the system of equations

$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}+2 y=0 \\
x+a y+a z-a=0
\end{array}\right.
$$

have a unique solution?
Answer: $a= \pm \frac{\sqrt{2}}{2}$.

Solution. Let's complete the square for $y$ in the first equation:

$$
x^{2}+(y+1)^{2}+z^{2}=1
$$

Now it is clear that this equation represents a sphere with center at the point $(0, -1, 0)$ and radius 1. The second equation represents a plane. The system of equations has a unique solution if the plane is tangent to the sphere.

This will be the case if the distance from the center of the sphere to the plane is equal to the radius of the sphere. Using the known formula for the distance from a point to a plane, we get

$$
\frac{|-a-a|}{\sqrt{1+a^{2}+a^{2}}}=1
$$

Then,

$$
|2 a|=\sqrt{1+2 a^{2}} ; \quad 4 a^{2}=1+2 a^{2} ; \quad a^{2}=\frac{1}{2}
$$

Thus, $a= \pm \frac{\sqrt{2}}{2}$.

Evaluation. 13 points for a correct solution.

## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

## Final Stage

$2017-2018$ academic year

## Problems, Answers, and Evaluation Criteria

## 11th Grade  Variant 2

## physics