Let $C_1, C_2$ and $C_3$ be three parallel chords of a circle on the same side of the center. 
The distance between $C_1$ and $C_2$ is the same as the distance between $C_2$ and $C_3$. 
The lengths of the chords are $20, 16$, and $8$. The radius of the circle is 
$\text{(A)} \ 12 \qquad  \text{(B)} \ 4\sqrt{7} \qquad  \text{(C)} \ \frac{5\sqrt{65}}{3} \qquad  \text{(D)}\ \frac{5\sqrt{22}}{2}\qquad \text{(E)}\ \text{not uniquely determined}$
Let the center of the circle be on the origin with equation $x^2 + y^2 = r^2$. 
As the chords are bisected by the x-axis their y-coordinates are $10, 8, 4$ respectively. Let the chord of length $10$ have x-coordinate $a$. Let $d$ be the common distance between chords. Thus, the coordinates of the top of the chords will be $(a, 10), (a+d, 8), (a+2d, 4)$ for the chords of length $20, 16$, and $8$ respectively. 
As these points fall of the circle, we get three equations: 
$\\a^2 + 100 = r^2$
$\\(a+d)^2 + 64 = r^2$
$\\(a+2d)^2 + 16 = r^2$
Subtracting the first equation from the second we get:
$\\(a+d)^2 - a^2 - 36 = 0$
$\\d(2a+d) = 36$
Similarly, by subtracting the first equation from the third we get:
$\\(a+2d)^2 - a^2 - 84 = 0$
$\\2d(2a+2d) = 84$
$\\d(a+d) = 21$
Subtracting these two equations gives us $ad = 15$. Expanding the second equation now gives us
$\\a^2 + 2ad + d^2 + 64 = r^2$
$\\a^2 + d^2 + 94 = r^2$
Subtracting the first equation from this yields:
$\\d^2 - 6 = 0$
$\\d = \sqrt{6}$
Combining this with $ad = 15$ we get
$\\\sqrt{6}a = 15$
$\\a = \frac{15}{\sqrt{6}} = \frac{5\sqrt{6}}{2}$
Plugging this into the first equation finally us
$\\(\frac{5\sqrt{6}}{2})^2 + 100 = r^2$
$\frac{150}{4} + 100 = \frac{550}{4} = r^2$
$\\r = \sqrt{\frac{550}{4}} = \frac{5\sqrt{22}}{2}$
$\fbox{D}$