Question

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 $a + b?$

$5\sqrt{2} + 4$

$\sqrt{17} + 7$

$6\sqrt{2} + 3$

$\sqrt{15} + 8$

$12$

Solution:

The ellipse $\dfrac{x^2}{16} + y^2 = 1$ has semi-axes $4$ and $1,$ so $c^2 = 16 - 1 = 15$ and the foci are $(\pm\sqrt{15}, 0).$

A circle through both foci has its center on the $y$ -axis, say $(0, k),$ with radius $\sqrt{k^2 + 15}.$ Its top point always lies outside the ellipse. For four intersection points, its bottom point $(0, k - \sqrt{k^2 + 15})$ must be below $y = -1,$ which happens exactly when $0 \le k \lt 7.$

As $k$ ranges over $[0, 7),$ the radius $\sqrt{k^2 + 15}$ ranges over $[\sqrt{15}, 8),$ so $a + b = \sqrt{15} + 8.$

Thus, the correct answer is D.

2015 AMC 12A Problem 21

Solution:

Problem 21 in Other Years