AMC12 2015 A

Answer (D): The ellipse with equation $x^2+16y^2=16$ is centered at the origin, with a major axis of length $8$ and a minor axis of length $2$. If the foci have coordinates $(\pm c,0)$, then $c^2+1^2=4^2$. Thus $c=\pm\sqrt{15}$. Any circle passing through both foci must have its center on the $y$-axis; thus $r$ is at least as large as the distance from the foci to the $y$-axis. That is, $r\ge\sqrt{15}$. For any $k\ge0$, the circle of radius $\sqrt{k^2+15}$ and center $(0,k)$ passes through both foci (in the interior of the ellipse) and the points $(0,k\pm\sqrt{k^2+15})$. The point $(0,k+\sqrt{k^2+15})$ is in the exterior of the ellipse since $k+\sqrt{k^2+15}>\sqrt{15}>1$. The point $(0,k-\sqrt{k^2+15})$ is in the exterior of the ellipse if and only if $k-\sqrt{k^2+15}<-1$, that is, if and only if $k<7$. Thus, for $k\ge0$, the circle with center $(0,k)$ intersects the ellipse in four points if and only if $0\le k<7$. As $k$ increases, the radius $r=\sqrt{k^2+15}$ increases as well, so the set of possible radii is the interval $[\sqrt{15},\sqrt{7^2+15})=[\sqrt{15},8)$. The requested answer is $\sqrt{15}+8$.