**A Proof**

**Given:**
Two fixed circles where one circle is in the interior of the other
and a tangent circle tangent to the larger fixed circle at a given
point such that the non-tangent points of the smaller given circle
are in the interior of the tangent circle.

**Prove:**
The locus of the centers of all possible tangent circles constructed
in this manner is an ellipse.

Consider the diagram below.

In order for the locus of the centers
of the tangent circles is an ellipse, the sum of the distances
from the two foci **A** and **B** to the center of the tangent
triangle **C** must be constant for any point of tangency **D**.
To summarize, **AC + BC** must be constant.

To show that the sum is constant, it
is helpful to first show that **Triangle EFC** is congruent
to **Triangle BFC**. This can be accomplished by asserting
that **Segments EF** and **BF** are congruent (due to **F**
being the midpoint of **Segment EB**), **Segment CF** being
congruent to itself (reflexive property) and **Angles EFC**
and **BFC** being congruent (since **Lines FC** and **EB**
are perpendicular, both angles are right angles and, therefore,
congruent). Thus, **Triangle EFC** is congruent to **Triangle
BFC** by SAS.

From these congruent triangles, corresponding
**Segments CB** and **CE** are congruent. Using substitution,
**AC + BC** becomes **AC + CE** which equals **AE**.

Is **AE** constant? Since **AE =
AD - DE** and both **AD** and **DE** represent the respective
radii of the fixed circles, it can be established that **AE**
is constant. Thus, for the possible points **D** on **Circle
A**, the locus of **C** is an ellipse. QED : )

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