$$
\begin{array}{l}
\text { 10. Given } f(x)=\left\{\begin{array}{ll}
-2 x, & x<0 ; \\
x^{2}-1, & x \geqslant 0,
\end{array}\right. \text { the equation } \\
f(x)+2 \sqrt{1-x^{2}}+\left|f(x)-2 \sqrt{1-x^{2}}\right|-2 a x-4=0
\end{array}
$$

has three real roots $x_{1}<x_{2}<x_{3}$. If $x_{3}-x_{2}=2\left(x_{2}-x_{1}\right)$, then the real number $a=$ $\qquad$
10. $\frac{\sqrt{17}-3}{2}$.

Let $g(x)=2 \sqrt{1-x^{2}}(-1 \leqslant x \leqslant 1)$.
By $\max \{f(x), g(x)\}$
$$
=\frac{1}{2}(f(x)+g(x)+|f(x)-g(x)|),
$$

the equation can be transformed into
$$
\begin{array}{l}
\max \{f(x), g(x)\}=a x+2 . \\
\text { By }-2 x \geqslant 2 \sqrt{1-x^{2}} \Rightarrow x \leqslant-\frac{\sqrt{2}}{2} .
\end{array}
$$

Then $\max \{f(x), g(x)\}$
$$
=\left\{\begin{array}{ll}
-2 x, & x \in\left[-1,-\frac{\sqrt{2}}{2}\right] ; \\
2 \sqrt{1-x^{2}}, & x \in\left[-\frac{\sqrt{2}}{2}, 1\right] .
\end{array}\right.
$$

Therefore, when $-2 x=a x+2$, solving gives
$x_{1}=-\frac{2}{a+2}\left(-1 \leqslant x_{1} \leqslant-\frac{\sqrt{2}}{2}\right)$
$\Rightarrow 0 \leqslant a \leqslant 2 \sqrt{2}-2$;
When $2 \sqrt{1-x^{2}}=a x+2$, solving gives
$x_{3}=0, x_{2}=-\frac{4 a}{a^{2}+4}<0$.
By $x_{3}-x_{2}=2\left(x_{2}-x_{1}\right) \Rightarrow 2 x_{1}=3 x_{2}$
$\Rightarrow \frac{4}{a+2}=\frac{12 a}{a^{2}+4} \Rightarrow a=\frac{\sqrt{17}-3}{2}$.