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

Since $2 \sqrt{1-x^{2}}=a x+2 \Rightarrow\left(a^{2}+4\right) x^{2}+4 a x=0$
$$
\Rightarrow x_{2}=-\frac{4 a}{a^{2}+4}, x_{3}=0 \text {, }
$$

Solving $\left\{\begin{array}{l}y=a x+2, \\ y=-2 x\end{array} \Rightarrow x_{1}=-\frac{2}{a+2}\right.$.
Thus $\frac{4 a}{a^{2}+4}=2\left(-\frac{4 a}{a^{2}+4}+\frac{2}{a+2}\right) \Rightarrow \frac{3 a}{a^{2}+4}=\frac{1}{a+2} \Rightarrow a^{2}+3 a-2=0 \Rightarrow a=\frac{-3+\sqrt{17}}{2}$.