Answer: E

Small Hint:

Place the chord intersection at the origin and the perpendicular chords on the coordinate axes

Big Hint:

The circle’s center lies on both chord perpendicular bisectors

Solution:

Place the intersection at $(0,0)$ with endpoints $(-3,0),$ $(4,0),$ $(0,-2),$ and $(0,6).$ The perpendicular bisectors of the chords are $x=\frac{1}{2}$ and $y=2,$ so the center is $(\frac{1}{2},2).$ Using the endpoint $(4,0),$ $$\begin{aligned} r^2 &=\left(4-\frac12\right)^2+(0-2)^2\\ &=\frac{65}{4}. \end{aligned}$$ Therefore the diameter is $2r=\sqrt{65}.$

Thus, the correct answer is E.