13 If the real numbers $x, y$ satisfy $x^{2}+y^{2}=1$, then the minimum value of $\frac{2 x y}{x+y-1}$ is $\qquad$

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13
$$
\begin{aligned}
\frac{2 x y}{x+y-1} & =\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{x+y-1} \\
& =\frac{(x+y)^{2}-1}{x+y-1} \\
& =x+y+1 .
\end{aligned}
$$

Since
$$
\left(\frac{x+y}{2}\right)^{2} \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2},
$$

we have
$$
-\sqrt{2} \leqslant x+y \leqslant \sqrt{2}
$$

Thus,
$$
\frac{2 x y}{x+y-1}=x+y+1 \geqslant-\sqrt{2}+1 \text {, }
$$

with equality if and only if $x=y=-\frac{\sqrt{2}}{2}$. Therefore, the minimum value of $\frac{2 x y}{x+y-1}$ is $1-\sqrt{2}$.