Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$. It is easy to see that $y = -\frac{x}{ab} + \frac{1}{a} + \frac{1}{b}$ is the equation of the line $AB$ and $y = abx$ is the equation of the line $OD$ because $OD$ is perpendicular to $AB$. Then we easily calculate the coordinates of $C$: $C = (\frac{1}{\sqrt{ab}}, \sqrt{ab})$.

Let $(x-t)^2 + (y-l)^2 = r^2$ be the equation of the circle $S$. Since $A, B, C$ lie on the hyperbola, we have
$$(x-t)^2 + \left(\frac{1}{x}-l\right)^2 = r^2 \implies x^4 - 2tx^3 + (t^2+l^2-r^2)x^2 - 2lx + 1 = 0. \quad (1)$$
Equation (1) has even number of real roots. Since $S$ passes through $A, B, C$, equation (1) has four real roots (not necessarily different). Let $\alpha$ be the fourth root of (1), then
$$1 = \alpha abc = \alpha \frac{ab}{\sqrt{ab}} = \alpha \sqrt{ab} \implies \alpha = \frac{1}{\sqrt{ab}}.$$
By Vieta's theorem,
$$2t = a + b + 2\frac{1}{\sqrt{ab}}, \quad 2l = \frac{1}{a} + \frac{1}{b} + 2\sqrt{ab}, \quad t^2 + l^2 - r^2 = ab + \frac{1}{ab} + 2\frac{a+b}{\sqrt{ab}}.$$
Hence
$$r^2 = \left(\frac{a+b}{2} + \frac{1}{\sqrt{ab}}\right)^2 + \left(\frac{a+b}{2ab} + \sqrt{ab}\right)^2 - ab - \frac{1}{ab} - 2\frac{a+b}{\sqrt{ab}} =$$
$$= \left(\frac{a+b}{2}\right)^2 \left(1+\frac{1}{a^2b^2}\right).$$

Let $a+b=n$, $\sqrt{ab}=m$, then
$$\left(x - \frac{n}{2} - \frac{1}{m}\right)^2 + \left(y - \frac{n}{2m^2} - m\right)^2 = \left(\frac{n}{2}\right)^2 \left(1 + \frac{1}{m^4}\right).$$
Find the coordinates of the intersection points of $S$ and the line $OD$. Since the equation of $OD$ is $y = abx = m^2x$, we have
$$\left(x - \frac{n}{2} - \frac{1}{m}\right)^2 + \left(m^2x - \frac{n}{2m^2} - m\right)^2 = r^2. \quad (2)$$
This equation has two roots, one of them is the abscissa of $C$, i.e., $\frac{1}{m}$ and the other is the abscissa of $F$ (it is easy to verify that $x = \frac{1}{m}$ satisfies (2)). Let $F = (f; g)$. Then, by Vieta's theorem,
$$\frac{1}{m} + f = \frac{2\left(\frac{n}{2} + \frac{1}{m} + \frac{n}{2} + m^3\right)}{1 + m^4} = \frac{2n}{m^4 + 1} + \frac{2}{m} \implies f = \frac{2n}{m^4 + 1} + \frac{1}{m}.$$
Then $g = abf = m^2f = m + \frac{2nm^2}{m^4 + 1}$. Therefore
$$CF^2 = \left(\frac{1}{m} - \frac{1}{m} - \frac{2n}{m^4 + 1}\right)^2 + \left(m - m - \frac{2nm^2}{m^4 + 1}\right)^2 = \\ = \left(\frac{2n}{m^4 + 1}\right)^2 (m^4 + 1) = \frac{4n^2}{m^4 + 1}.$$
Let $D = (d; e)$. Since $D$ is the intersection point of $OD$ and $AB$, we have
$$abd = -\frac{d}{ab} + \frac{1}{a} + \frac{1}{b} \implies \frac{a+b}{ab} = d \left(ab + \frac{1}{ab}\right) \implies$$
$$d = \frac{a+b}{1+a^2b^2} = \frac{n}{m^4+1},$$
$$e = abd = m^2d = \frac{m^2n}{m^4 + 1}.$$
Therefore
$$OD^2 = \frac{n^2}{(m^4 + 1)^2} + \frac{m^4 n^2}{(m^4 + 1)^2} = \frac{n^2}{m^4 + 1}.$$
Then
$$\frac{OD^2}{CF^2} = \frac{n^2}{m^4 + 1} \cdot \frac{m^4 + 1}{4n^2} = \frac{1}{4} \implies \frac{OD}{CF} = \frac{1}{2}.$$