8. The function $f(x)=\frac{\sqrt{a^{2}-x^{2}}}{|x+a|-a}$ is an odd function if and only if $\qquad$ .
8. $a>0$. Hint: Necessity. If $f(x)=\frac{\sqrt{a^{2}-x^{2}}}{|x+a|-a}$ is an odd function, then $a \neq 0$ (otherwise, the domain of $f(x)$ would be an empty set), and from $\frac{\sqrt{a^{2}-x^{2}}}{|-x+a|-a}=\frac{-\sqrt{a^{2}-x^{2}}}{|x+a|-a}$ we get $|x+a|+|x-a|=2 a, \therefore a>0$.

Sufficiency. If $a>0$, then the domain of $f(x)$ is $[-a, 0) \cup(0, a]$, at this time $f(x)=\frac{\sqrt{a^{2}-x^{2}}}{x}$, clearly $f(-x)=$ $-f(x), f(x)$ is an odd function.