# Line Joining Centers of Two Circles Touching Externally

## Theorem

In the words of Euclid:

*If two circles touch one another externally, the straight line joining their centres will pass through the point of contact.*

(*The Elements*: Book $\text{III}$: Proposition $12$)

## 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 $A$.

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:

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.

$\blacksquare$

## Historical Note

This proof is Proposition $12$ of Book $\text{III}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions