A circle with center $(0,k)$, $k>8$, is tangent to the lines $y=2x$, $y=-2x$, and $y=8$. What is the radius of this circle?

1. Let the radius be $r$. The circle is tangent to $y=8$ and lies above it, with center $D=(0,k)$, so $r=k-8$.
2. The distance from $D=(0,k)$ to the line $2x-y=0$ is $\dfrac{|-k|}{\sqrt{2^2+1^2}}=\dfrac k{\sqrt5}$, and this equals $r$ (by symmetry the same holds for $y=-2x$).
3. So $\dfrac k{\sqrt5}=k-8$, giving $k\left(1-\dfrac1{\sqrt5}\right)=8$, so $k=\dfrac{8\sqrt5}{\sqrt5-1}=2\sqrt5(\sqrt5+1)=10+2\sqrt5$.
4. The radius is $r=k-8=\boxed{2+2\sqrt5}\approx6.47$.