Line Joining Centers of Two Circles Touching Externally

Theorem
Let two circles touch externally.

Then the straight line joining their centers passes through the point where they touch.

Proof
Let the circles $$ABC$$ and $$ADE$$ touch externally at $$A$$.

Let $$F$$ be the center of $$ABC$$ and let $$G$$ be the center of $$ADE$$.

We are to show that the straight line joining $$F$$ to $$G$$ passes through $$A$$.


 * Euclid-III-12.png

Suppose, as in the diagram above, that it does not, and that it were possible for it to pass through $$C$$ and $$D$$, as $$FCDG$$.

(It is clear that the diagram does not have $$F$$ and $$G$$ as the actual centers of these circles - it is the point of this proof to demonstrate that this would not be possible.)

Join $$AF$$ and $$AG$$.

We have that:
 * Since $$F$$ is the center of $$ABC$$, then $$FA$$ and $$FC$$ are both radii of $$ABC$$, and so $$FA = FC$$.


 * Since $$G$$ is the center of $$ADE$$, then $$GA$$ and $$GD$$ are both radii of $$ADE$$, and so $$GA = GD$$.

So $$FA + AG = FC + GD$$.

So all of $$FCDG$$ is greater than $$FA + AG$$.

But from Sum of Two Sides of Triangle Greater than Third Side $$FA + AG$$ is greater than $$FCDG$$.

Hence we have a contradiction, and so $$FG$$ has to go through point $$A$$, the point of contact of the two circles.

Hence the result.