8. Given $x, y, z \in \mathbf{R}_{+}$, and $\sqrt{x^{2}+y^{2}}+z=1$. Then the maximum value of $x y+2 x z$ is $\qquad$ .
8. $\frac{\sqrt{3}}{3}$.
$$
\begin{array}{l}
\text { Given } \sqrt{x^{2}+y^{2}}+z=1 \\
\Rightarrow x^{2}+y^{2}=(1-z)^{2} \\
\Rightarrow x^{2}=(1-z-y)(1-z+y) .
\end{array}
$$

Then $(x y+2 x z)^{2}=x^{2}(y+2 z)^{2}$
$$
\begin{array}{l}
=\frac{1}{3}(3-3 z-3 y)(1-z+y)(y+2 z)(y+2 z) \\
\leqslant \frac{1}{3}\left[\frac{(3-3 z-3 y)+(1-z+y)+(y+2 z)+(y+2 z)}{4}\right]^{4} \\
=\frac{1}{3} .
\end{array}
$$

Therefore, $x y+2 x z \leqslant \frac{\sqrt{3}}{3}$, and the maximum value $\frac{\sqrt{3}}{3}$ is achieved when $x=\frac{\sqrt{3}}{3}, y=z=\frac{1}{3}$.