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I have the Dandelin sphere construction. That is, I am given a vertical cylinder with radius $r$ and two spheres of radius $r$ are put inside of it. A plane (horizontal or otherwise, just not vertical) goes through the cylinder and the two spheres are placed such that they are just tangent to the plane at points $F_1$ and $F_2$.

So, define a point $P$ to be any point in the intersection between the plane and the (hollow) cylinder. We have that $|F_1 - F_2| = 2c$ and $P - F_1| + |P - F_2| = 2a$. I want to show that $r = \sqrt{a^2 - c^2}$.

I've been trying to work something out using the pythagorean theorem, but I can't get something started.

MT_
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1 Answers1

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I'd never heard of Dandelin spheres, so I learned something finding this answer. In the figure below (apologies for the image quality, and all the extra labels, etc.), I'll show one thing (with a proof blithely adapted (i.e., stolen) from Wikipedia). The key fact I'll use (leaving the proof to you) is that if $PQ$ and $PR$ are segments tangent to a sphere $S$ at points $Q$ and $R$, then their lengths are equal.

Main claim: $|E_1E_2| = 2a$, so the spheres are separated by distance $2a$ between their centers.

Proof: The segments $PF_1$ and $PF_2$ are both tangent to the upper sphere, so their lengths are the same number $u$. Similarly for $PF_2$ and $PE_2$, with equal lengths $v$. Thus the distance $|E_1E_2|$ is $u + v = 2a$.
enter image description here

It then follows that the midpoint $Q$ of $F_1F_2$ is at distance $a$ from the center $A$ of the lower sphere. Since the length of $F_1F_2$ is $2c$, the distance from $Q$ to $F_2$ is $c$. Now $AF_2Q$ is a right triangle, with legs $r$ and $c$ and hypotenuse $a$, so $r^2 + c^2 = a^2$, hence $r = \sqrt{a^2 - c^2}$.

John Hughes
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