Circle Intercepting An Ellipse A circle of radius $r$ passes through both foci of, and exactly four points on, the ellipse with equation $x^2+16y^2=16$. The set of all possible values of $r$ is an interval $[a,b)$. What is $a+b$?
To find the value of the radius of the circle is quite easy. This is just half the distance from the two foci of the ellipse which are $(\sqrt{-15},0)$ and $(\sqrt{15},0)$, or $\sqrt{15}$.
I have no way to find $b$. Please help.
Note: I'm still in high school, so don't expect much from me.
 A: Just for fun, here is an animation of the family of circles that satisfy the criteria of the problem:

For those with Mathematica, an interactive form of this animation can be produced with the following command:
Manipulate[
 Show[
  ParametricPlot[
   {{4 Cos[t], Sin[t]}, Sqrt[d^2 + 15] {Cos[t], Sin[t]} + {0, d}},
   {t, 0, 2 Pi}, AspectRatio -> Automatic, PlotRange -> {{-8, 8}, {-8, 8}}],
  Graphics[{Red, Disk[#, .1] & /@ ({#[[1]] 4 Sqrt[2] Sqrt[105 - d^2
      + #[[2]] d Sqrt[15 + d^2]], -d + #[[2]] Sqrt[15 + d^2]}/15
      & /@ Tuples[{-1, 1}, 2]), Blue, Disk[{Sqrt[15], 0}, .05],
      Disk[{-Sqrt[15], 0}, .05], Orange, Disk[{0, d}, 0.1],
      Opacity[0.5], Line[{{Sqrt[15], 0}, {0, d}, {-Sqrt[15], 0}}]}]],
  {d, -7, 7}]

A: First find the focii of the ellipse, which are $(-\sqrt{15}, 0)$ and $(\sqrt{15}, 0)$


*

*Minimum radius will be when circle's centre is at $(0,0)$, that is, $\min(r) = \sqrt{15}$,
This is minimum as the focal points lie diametrically opposite to each other on circle.

*For maximum, consider the other boundary line condition, as said by JM.
If the radius is so large that the circle passing through focii becomes tangent to ellipse, then by pythagorous theorem:
$$(r-1)^2 + 15 = r^2$$
From which we get $r = 8$

PS: Dude I too am in high school. Cheers!
