Answer: 65

Small Hint:

Write the diameter as $10a+b$ and the chord as $10b+a$

Big Hint:

The squared distance from the center to the chord is one fourth the difference of their squares

Solution:

Let the diameter be $10a+b$ and the chord be $10b+a.$ Since a perpendicular from the center bisects a chord, $$\begin{aligned} 4OH^2&=(10a+b)^2\\ &\quad-(10b+a)^2\\ &=99(a^2-b^2). \end{aligned}$$ For $OH$ to be rational, $11(a^2-b^2)$ must be a square. Thus it equals $121c^2,$ so $$(a-b)(a+b)=11c^2.$$ Because $a$ and $b$ are digits with $a>b,$ the left side is at most $81,$ so $c=1$ or $2.$ If $c=1,$ the factors are $1$ and $11,$ giving $a=6$ and $b=5.$ If $c=2,$ the factor pairs of $44$ either have different parity or give $a>9.$ Hence the diameter is $65.$