11.5. Find all values of the parameter $a \in R$, for which the system

$$
\left\{\begin{array}{c}
\sqrt{x^{2}+y^{2}+6 x+10 y+34}+\sqrt{x^{2}+y^{2}+6 x+6 y+18}=2 \\
y^{2}+x^{2}-4 x+10=a
\end{array}\right.
$$

has a unique solution in $R \times R$.
Solution. We will solve this system geometrically. The solutions of the system are the points of intersection of the plane curves determined by the equations of the system. The initial system is equivalent to the following system:

$$
\left\{\begin{array}{c}
\sqrt{(x+3)^{2}+(y+5)^{2}}+\sqrt{(x+3)^{2}+(y+3)^{2}}=2 \\
(x-2)^{2}+y^{2}=a-6
\end{array} \Rightarrow a-6 \geq 0\right.
$$

The second equation of the system determines a circle with center at point $O(2 ; 0)$ and radius $R=\sqrt{a-6}$ for $a>6$ or a point $O(2 ; 0)$ for $a=6$. This circle has a fixed center and an arbitrary radius, which depends on the parameter $a$. It is necessary to find such values of the parameter $a$ that this circle intersects the graph determined by the first equation at only one point.

The distance between two points $A\left(x_{1} ; y_{1}\right)$ and $B\left(x_{2} ; y_{2}\right)$ can be found according to the formula

$$
d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}
$$

According to the distance formula, the first equation of the system shows that the sum of the distances from point $M(x ; y)$ to points $A(-3 ;-5)$ and $B(-3 ;-3)$ is equal to 2. Since the distance between points $A$ and $B$ is equal to 2, it follows that point $M(x ; y) \in[A B]$. Therefore, the geometric locus of points determined by the first equation of the system is the segment $[A B]$. According to the hypothesis, the following relationships must hold:

$$
\begin{aligned}
O B & \leq R \leq O A \\
O B^{2} & \leq R^{2} \leq O A^{2} \\
5^{2}+3^{2} & \leq a-6 \leq 5^{2}+5^{2} \\
40 & \leq a \leq 56 .
\end{aligned}
$$

Thus, $a \in[40 ; 56]$.