8. (2003 Hunan Province High School Mathematics Competition) Let $\min \{a, b\}$ be the minimum of two numbers $a, b$. When positive numbers $x, y$ vary, $t=\min \left\{x, \frac{y}{x^{2}+y^{2}}\right\}$ also varies. Then the maximum value of $t$ is $\qquad$.
8. If $x \leqslant \frac{y}{x^{2}+y^{2}}$, then $t=x \cdot t^{2}=x^{2} \leqslant x \cdot \frac{y}{x^{2}+y^{2}} \leqslant \frac{x y}{2 x y}=\frac{1}{2}$, when $x \geqslant \frac{y}{x^{2}+y^{2}}$, then $t^{2}=\left(\frac{y}{x^{2}+y^{2}}\right)^{2}$ $\leqslant \frac{x y}{x^{2}+y^{2}} \leqslant \frac{1}{2}$, hence $t \leqslant \frac{\sqrt{2}}{2}$, equality holds if and only if $x=y=\frac{\sqrt{2}}{2}$.
Therefore, the maximum value of $t$ is $\frac{\sqrt{2}}{2}$.