1. Let $x^{2}+y^{2}-2 x-2 y+1=0(x, y \in$
$\mathbf{R})$. Then the minimum value of $F(x, y)=\frac{x+1}{y}$ is $\qquad$ .
II. $1 . \frac{3}{4}$.
The original expression is transformed into $(x-1)^{2}+(y-1)^{2}=1$. Let $x-1=\cos \theta, y-1=\sin \theta$, then $x=1+\cos \theta, y=1+\sin \theta$. Therefore, $\frac{x+1}{y}=\frac{2+\cos \theta}{1+\sin \theta}$, let this be $\mu$. Then $\mu+\mu \sin \theta=2+\cos \theta$, which means $\mu \sin \theta-\cos \theta=2-\mu$. So,
$$
\sqrt{\mu^{2}+1} \sin (\theta-\varphi)=2-\mu .
$$
(Where, $\left.\cos \varphi=\frac{\mu}{\sqrt{\mu^{2}+1}}, \sin \varphi=\frac{1}{\sqrt{\mu^{2}+1}}\right)$
$$
\sin (\theta-\varphi)=\frac{2-\mu}{\sqrt{\mu^{2}}}+\left|\frac{2-\mu}{\sqrt{\mu^{2}}+1}\right| \leqslant 1 .
$$

Solving this, we get $\mu \geqslant \frac{3}{4}$, so the minimum value of $F(x, y)$ is $\frac{3}{4}$.