10. (20 points) Given the functions
$$
f(x)=x^{2}-2 a x \text { and } g(x)=-x^{2}-1
$$

their graphs have two common tangents, and the perimeter of the quadrilateral formed by these four tangent points is 6. Find the value of the real number $a$.
10. Let the functions $f(x)$ and $g(x)$ have a common tangent line passing through the points $\left(x_{1}, f\left(x_{1}\right)\right)$ and $\left(x_{2}, g\left(x_{2}\right)\right)$. Then the equation of the common tangent line is
$$
\begin{array}{l}
y=f\left(x_{1}\right)+f^{\prime}\left(x_{1}\right)\left(x-x_{1}\right) \\
\equiv g\left(x_{2}\right)+g^{\prime}\left(x_{2}\right)\left(x-x_{2}\right) . \\
\text { Hence } f^{\prime}\left(x_{1}\right)=g^{\prime}\left(x_{2}\right), \text { and } \\
f\left(x_{1}\right)-f^{\prime}\left(x_{1}\right) x_{1}=g\left(x_{2}\right)-g^{\prime}\left(x_{2}\right) x_{2} .
\end{array}
$$

Thus, $f^{\prime}\left(x_{1}\right)=g^{\prime}\left(x_{2}\right)$, and

Notice that,
$$
\begin{array}{l}
f^{\prime}(x)=2 x-2 a, g^{\prime}(x)=-2 x \\
\Rightarrow x_{1}+x_{2}=a, x_{1}^{2}+x_{2}^{2}=1 \\
\Rightarrow x_{1} x_{2}=\frac{a^{2}-1}{2} .
\end{array}
$$

Therefore, $x_{1}$ and $x_{2}$ are the two real roots of the equation $x^{2}-a x+\frac{a^{2}-1}{2}=0$.

Since $f(x)$ and $g(x)$ have two common tangents, $f(x)$ and $g(x)$ do not intersect.
Thus, $x_{1} \neq x_{2}$.
By $\Delta=a^{2}-4 \cdot \frac{a^{2}-1}{2}>0 \Rightarrow a^{2}<2$.
Let the coordinates of the four points be
$$
\begin{array}{l}
M\left(x_{1}, f\left(x_{1}\right)\right), N\left(x_{2}, f\left(x_{2}\right)\right), \\
P\left(x_{2}, g\left(x_{2}\right)\right), Q\left(x_{1}, g\left(x_{1}\right)\right) .
\end{array}
$$

Then $|P Q|^{2}=\left(x_{1}-x_{2}\right)^{2}+\left(g\left(x_{1}\right)-g\left(x_{2}\right)\right)^{2}$
$$
\begin{array}{l}
=\left(x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}\right)\left[1+\left(x_{1}+x_{2}\right)^{2}\right] \\
=\left(2-a^{2}\right)\left(1+a^{2}\right), \\
|M Q|=\left|f\left(x_{1}\right)-g\left(x_{1}\right)\right| \\
=\left|x_{1}^{2}-2 a x_{1}+x_{1}^{2}+1\right|=2-a^{2} .
\end{array}
$$

Similarly, $|M N|^{2}=\left(2-a^{2}\right)\left(1+a^{2}\right)$,
$$
|N P|=2-a^{2} \text {. }
$$

Therefore, the quadrilateral $M N P Q$ is a parallelogram, and
$2\left[\sqrt{\left(2-a^{2}\right)\left(1+a^{2}\right)}+2-a^{2}\right]=6$.
Solving for $a$ gives $a= \pm \frac{\sqrt{2}}{2}$, which is the desired result.