jmc

Question

An ellipse is drawn with major and minor axes of lengths 10 and 8 respectively. Using one focus as a center, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. Compute the radius of the circle.

Accepted Answer

Place the ellipse in the coordinate plane, as usual, so that the center is at the origin. Then the equation of the ellipse is

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1.

Also, the distance from the center to each foci is

5^{2} - 4^{2} = 3,

so one foci is at

F = (3, 0) .

Consider the circle centered at

F

with radius 2. The equation of this circle is

(x - 3)^{2} + y^{2} = 4,

so

y^{2} = 4 - (x - 3)^{2} .

Substituting into the equation of the ellipse, we get

\frac{x ^{2}}{25} + \frac{4 - ( x - 3 ) ^{2}}{16} = 1.

This simplifies to

3 x^{2} - 50 x + 175 = 0,

which factors as

(x - 5) (3 x - 35) = 0.

The solutions are

x = 5

and

x = \frac{35}{3},

the latter root being extraneous. This tells us that the ellipse and circle intersect only at the point

(5, 0),

and clearly we cannot draw a larger circle.

Hence, the maximum radius is

2 .