71 If real numbers $x, y$ satisfy $x^{2}+y^{2}=1$, then the minimum value of $\frac{2 x y}{x+y-1}$ is
A. $1-\sqrt{2}$
B. $1-2 \sqrt{2}$
C. 1
D. $-\sqrt{2}$
71 A. Let $x=\sin \theta, y=\cos \theta$, then
$$
\begin{aligned}
\frac{2 x y}{x+y-1} & =\frac{2 \sin \theta \cos \theta}{\sin \theta+\cos \theta-1}=\frac{(\sin \theta+\cos \theta)^{2}-1}{\sin \theta+\cos \theta-1} \\
& =\sin \theta+\cos \theta+1=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right)+1
\end{aligned}
$$

Therefore, the correct choice is A.