Twelve, as shown in Figure 5, the length of the diameter AB of a circle is a two-digit integer (in decimal). By reversing the two digits, we get the length of the chord CD, which is perpendicular to the diameter AB. The distance from the intersection point H to the center O of the circle is a positive rational number. Find the length of AB.
Let $\mathrm{AB}=10 \mathrm{x}+\mathrm{y}$, where $\mathrm{x}, \mathrm{y}=0,1,2, \cdots, 9$, and $x \neq 0$.
Then $C D=10 y+x$.
$$
\begin{aligned}
\mathrm{OH}^{2} & =\left(\frac{10 \mathrm{x}+\mathrm{y}}{2}\right)^{2}-\left(\frac{10 \mathrm{y}+\mathrm{x}}{2}\right)^{2} \\
& =\frac{9}{4} \cdot 11\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) .
\end{aligned}
$$

Since $0 H$ is a rational number, then $11\left(x^{2}-y^{2}\right)$ must be a perfect square. Let
$$
11\left(x^{2}-y^{2}\right)=(11 k)^{2},
$$

then $x^{2}-y^{2}=(x+y)(x-y)=11 k^{2}$.
Obviously $x+y \geqslant x-y$, and $0<x+y \leqslant 18$, $0<x-y<8$,
thus we can only have $x+y=11, x-y=1$.
Solving gives $\mathrm{x}=6, \mathrm{y}=5$,
i.e., $\mathrm{AB}=65$.