取曲线 $y = \dfrac{1}{x - 1}$ 上的点 $\left(a,\dfrac{1}{a - 1}\right)$，$a\neq1$，因为圆心 $ \left(0,1\right) $ 与点 $\left(a,\dfrac{1}{a - 1}\right)$ 之间的距离$$ \begin{split}
d &=\sqrt{a^{2} +\left(\dfrac{1}{a - 1} - 1\right)^{2}}\\
&=\sqrt{\left(a-1\right)^2+2\left(a-1\right)+\dfrac{1}{\left(a-1\right)^2}-\dfrac{2}{a-1}+2}\\
&=\sqrt{\left(a-1-\dfrac{1}{a-1}\right)^2+2\left(a-1-\dfrac{1}{a-1}\right)+4}\\
&=\sqrt{\left(a-1-\dfrac{1}{a-1}+1\right)^2+3}\\
&\geqslant\sqrt{3} .\
\end{split} $$上式中当且仅当 $a=\dfrac {1\pm \sqrt 5}{2}$ 时取得等号．
所以，圆 $x^{2} + \left(y - 1\right)^{2} = r^{2} $ 与曲线 $\left(x - 1\right)y = 1$ 没有公共点，则半径 $r$ 的取值范围是 $\left(0,\sqrt 3\right)$．