2015 AMC 12A Problems/Problem 21

Problem

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 $a+b?$

$\textbf{(A)}\ 5\sqrt{2}+4\qquad\textbf{(B)}\ \sqrt{17}+7\qquad\textbf{(C)}\ 6\sqrt{2}+3\qquad\textbf{(D)}}\ \sqrt{15}+8\qquad\textbf{(E)}\ 12$ (Error compiling LaTeX. Unknown error_msg)

Solution

We can graph the ellipse by setting $x = 0$ and finding possible values for $y$ , and vice versa. The points where the ellipse intersects the coordinate axes are $(0, 1), (0, -1), (4, 0)$ , and $(-4, 0)$ . Recall that the two foci lie on the major axis of the ellipse and are a distance of $c$ away from the center of the ellipse, where $c^2 = a^2 - b^2$ , with $a$ being the length of the major (longer) axis and $b$ being the minor (shorter) axis of the ellipse. We have that $c^2 = 4^2 - 1^2 \implies$ $c^2 = 15 \implies c = \pm \sqrt{15}$ . Hence, the coordinates of both of our foci are $(0, \sqrt{15})$ and $(0, -\sqrt{15})$ . In order for a circle to pass through both of these foci, we must have that the center of this circle lies on the y-axis.

The minimum possible value of $r$ belongs to the circle whose diameter's endpoints are the foci of this ellipse, so $a = \sqrt{15}$ . The value for $b$ is achieved when the circle passes through the foci and only three points on the ellipse, which is possible when the circle touches $(0, 1)$ or $(0, -1)$ . Which point we use does not change what value of $b$ is attained, so we use $(0, -1)$ . Here, we must find the point $(0, y)$ such that the distance from $(0, y)$ to both foci and $(-1, 0)$ is the same. Now, we have the two following equations. $(\sqrt{15})^2 + (y)^2 = b^2$ $y + 1 = b \implies y = b - 1$ Substituting for $y$ , we have that $15 + (b - 1)^2 = b^2 \implies -2b + 16 = 0.$

Solving the above simply yields that $b = 8$ , so our answer is $a + b = \sqrt{15} + 8 \textbf{ (D)}$ .

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2015 AMC 12A Problems/Problem 21

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Solution