smc

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 $(\frac{5\sqrt{6}}{2}{)}^2+100=r^2$ $\frac{150}{4}+100=\frac{550}{4}=r^2$ $r=\sqrt{\frac{550}{4}}=\frac{5\sqrt{22}}{2}$ $\boxed{\frac{5\sqrt{22}}{2}}$