# Solution:

Since $B F$ and $A F$ are segments of tangents, $B F=A F=2$.

Right triangles $A O C$ and $B F C$ are similar, $\frac{O C}{F C}=\frac{A C}{B C}=\frac{O A}{B F}$.

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Let $x=C F, \quad y=B C$.

Then $\frac{y+2 \sqrt{5}}{x}=\frac{2+x}{y}=\sqrt{5} \Leftrightarrow\left\{\begin{array}{l}y+2 \sqrt{5}=x \sqrt{5}, \\ 2+x=y \sqrt{5},\end{array} \Leftrightarrow\left\{\begin{array}{l}\sqrt{5} y+10=5 x, \\ 2+x=\sqrt{5} y,\end{array} \Leftrightarrow \Leftrightarrow\left\{\begin{array}{l}x=3, \\ y=\sqrt{5} .\end{array}\right.\right.\right.$

In triangle $A B C$, we have $A C=5, \quad B C=\sqrt{5}, \sin \angle O C A=\frac{A O}{O C}=\frac{2 \sqrt{5}}{3 \sqrt{5}}=\frac{2}{3}, \cos \angle O C A=\frac{\sqrt{5}}{3}$.

From this, by the cosine rule, we get

$$A B^{2}=A C^{2}+B C^{2}-2 A C \cdot B C \cos \angle O C A=25+5-10 \cdot 5 \cdot \frac{1}{3}=\frac{40}{3}, A B=\frac{2 \sqrt{10}}{\sqrt{3}}$$

Then $R_{\text {circum }}=\frac{A B}{2 \sin \angle O C A}=\frac{2 \sqrt{10} \cdot 3}{2 \sqrt{3} \cdot 2}=\frac{\sqrt{30}}{2}$.

Answer: $\sqrt{30} / 2$.