(14 Given real numbers $a$, $b$, $c$, $d$ satisfy $ab=c^2+d^2=1$, then $(a-c)^2+$ $(b-d)^2$ the minimum value is $\qquad$ .
(14) $3-2 \sqrt{2}$ Hint: The minimum value of $(a-c)^{2}+(b-d)^{2}$ is the square of the minimum distance between a point on the hyperbola $ab=1$ and a point on the circle $c^{2}+d^{2}=1$, which is the square of the distance between the points $(1,1)$ and $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, i.e., $3-2 \sqrt{2}$.