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