10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution

$$
\left\{\begin{array}{l}
2(x+7)^{2}+(y-4)^{2}=c \\
(x+4)^{2}+2(y-7)^{2}=c
\end{array}\right.
$$
Answer: $c=6.0$.

Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have

$$
\begin{aligned}
& \left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right) \geqslant(|x+\alpha|+|y-\beta|)^{2} \\
& \left(1+\frac{1}{2}\right)\left((x+\beta)^{2}+2(y-\alpha)^{2}\right) \geqslant(|x+\beta|+|y-\alpha|)^{2}
\end{aligned}
$$

Therefore, for any solution $(x, y)$ of the system, we have

(3) $3 c \geqslant(|x+\alpha|+|y-\beta|)^{2}+(|x+\beta|+|y-\alpha|)^{2} \geqslant$

$$
\begin{aligned}
\geqslant(\alpha-\beta+x+y)^{2}+(\alpha-\beta-(x+y))^{2} & = \\
& =2(\alpha-\beta)^{2}+2(x+y)^{2} \geqslant 2(\alpha-\beta)^{2}
\end{aligned}
$$

Thus, if $c<\frac{2}{3}(\alpha-\beta)^{2}$, the system has no solutions.

If $c=\frac{2}{3}(\alpha-\beta)^{2}$, then equality must be achieved in all inequalities (1)-(3). Therefore, $x+y=0$, the numbers $x+\alpha$ and $x+\beta$ must have different signs, and

$$
\frac{2(x+\alpha)^{2}}{(x+\beta)^{2}}=\frac{1}{2}
$$

From these conditions, it follows that when $c=\frac{2}{3}(\alpha-\beta)^{2}$, there is only one solution $\left(-\frac{2 \alpha+\beta}{3}, \frac{2 \alpha+\beta}{3}\right)$. Remark. One can immediately notice that the solution of the system is symmetric with respect to the line $y=-x$ (two ellipses with centers at points $(-\alpha ; \beta)$ and $(-\beta ; \alpha)$). The solution will be unique if one of the ellipses touches this line. This is satisfied if the equation $2(x+\alpha)^{2}+(x+\beta)^{2}=c$ has a unique solution. Setting the discriminant to zero, we find the solution. https://ggbm.at/FVmUGS4Y