Let ellipse $4x^2+16y^2=64$ and circle $x^2+y^2=r^2$ have a common tangent touching at A and B respectively. The maximum length of AB can be?
(A) 4 (B) 3 (C) 2 (D) 5
I tried solving the question using parametric coordinates, using slope form of tangents but there are a lot more variables and much-complicated equations. Can't think of some intuitive or out-of-the-box solution. Please help.
The dual conics are $16X^2+4Y^2=1$ and $r^2X^2+r^2Y^2=1,$ and they intersect when r is between $2$ and $4$ making the common tangents $$\pm x \sqrt{r^2-4}/(2 \sqrt3 r) \pm y \sqrt{16-r^2}/(2\sqrt3 r) + 1=0$$ and $$A: (\pm(r \sqrt{r^2-4})/(2 \sqrt3),\pm(r \sqrt{16-r^2})/(2 \sqrt3))$$ $$B: ((\pm 8\sqrt{r^2-4})/(\sqrt3 r),\pm (2 \sqrt{16-r^2})/(\sqrt3 r)),$$ so the squared distance is $((4-r)(4+r)(r-2)(r+2))/r^2$ which maximizes at $r=\sqrt8$ when the distance is $$2.$$