Line Joining Centers of Two Circles Touching Externally

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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 $A$.


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.

$\blacksquare$


Historical Note

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


Sources