14. Let real numbers $a, b, c, d$ satisfy $ab=c^{2}+d^{2}=1$. Then the minimum value of $(a-c)^{2}+(b-d)^{2}$ is $\qquad$
14. $3-2 \sqrt{2}$.

It is easy to see that 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., $3-2 \sqrt{2}$.