smc

Let the chords intersect the circle at points $A,B,C,$ and $D$ to form simple polygon $ABCD$. Further, let the chords intersect at point $P$ with $AP=6,PC=2,BP=3,$ and $PD=4$, as in the diagram. Then, because $\overline{AC}\perp \overline{BD}$, by the Pythagorean Theorem, $AB=3\sqrt{5}$ and $BC=\sqrt{13}$. Because a circle is determined by three coplanar points, the circumcircle of $△ABC$ will be the circumcircle of $ABCD$, so the circumdiameter of $△ABC$ will be our desired answer. We know that $[△ABC]=\frac{(6+2)\cdot 3}{2}=12$. Furthermore, we know that we can express this area as $\frac{abc}{4R}$, where $a,b,$ and $c$ are $△ABC$'s side lengths and $R$ is its circumradius. Setting this expression equal to $12$, we can now solve for $R$: $$\begin{array}{rl}\frac{abc}{4R}& =12\\ \frac{\sqrt{13}\cdot 8\cdot 3\sqrt{15}}{4R}& =12\\ \frac{2\cdot 3\sqrt{65}}{R}& =12\\ 6\sqrt{65}& =12R\\ R& =\frac{\sqrt{65}}{2}\end{array}$$ Because the problem asks for the diameter of the circle, our answer is $2R=2\cdot \frac{\sqrt{65}}{2}=\boxed{\sqrt{65}}$.