13. If real numbers $x, y$ satisfy $x^{2}+y^{2}=1$, then the minimum value of $\frac{2 x y}{x+y-1}$ is $\qquad$
$13.1-\sqrt{2}$.
Notice
$$
\begin{array}{l}
\frac{2 x y}{x+y-1}=\frac{(x+y)^{2}-\left(x^{2}+y^{2}\right)}{x+y-1} \\
=\frac{(x+y)^{2}-1}{x+y-1}=x+y+1 .
\end{array}
$$

Since $\left(\frac{x+y}{2}\right)^{2} \leqslant \frac{x^{2}+y^{2}}{2}=\frac{1}{2}$, therefore, $-\sqrt{2} \leqslant x+y \leqslant \sqrt{2}$.
Thus, $\frac{2 x y}{x+y-1}=x+y+1 \geqslant-\sqrt{2}+1$, equality holds if and only if $x=y=-\frac{\sqrt{2}}{2}$.
Hence, the minimum value of $\frac{2 x y}{x+y-1}$ is $1-\sqrt{2}$.