11.7 Find all values of the parameter \(a\), for each of which the system of equations 
\[
\left\{
\begin{array}{l}
x^{2}+y^{2}=2 a, \\ 
x+\log _{2}\left(y^{2}+1\right)=a
\end{array}
\right.
\]
has a unique solution.
Answer: $a=0$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7a0ab87e7d282e4dcdc0g-1.jpg?height=636&width=671&top_left_y=1241&top_left_x=378)

Solution. Note that if $(x, y)$ is a solution to the system, then $(x, -y)$ is also a solution to this system. Therefore, if for a parameter $a$ the system has a unique solution, the value of $y$ must be zero. Then from the second equation, we have $x=a$, and substituting $x=a$, $y=0$ into the first equation, we get $a^2 = 2a$. Thus, $a=0$ or $a=2$. Now we need to check these values of $a$ (note that they are derived as a consequence of the uniqueness of the solution to the system, but do not yet guarantee uniqueness).

For $a=0$, the first equation is equivalent to the equalities $x=y=0$, and for these zero values of the unknowns, the second equation is satisfied, i.e., the value $a=0$ works.

For $a=2$, the system has the solution $x=2, y=0$, but for the system, this solution is not unique. Indeed, the curve $x=2-\log_2(y^2+1)$ intersects the circle $x^2+y^2=4$ not only at the point $(2; 0)$ but also at two symmetric points in the second and third quadrants. (see figure). This follows from the fact that the points $(1; 1)$ and $(1; -1)$ lie on the given curve and are inside the circle $x^2+y^2=4$ (or one can verify that the points $(0; \pm \sqrt{3})$ on the y-axis will also be such points), and for $|y|>2$, the corresponding points on the curve are outside the circle.