Circles Touch at One Point at Most

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Theorem

In the words of Euclid:

A circle does not touch a circle at more points than one, whether it touch it internally or externally.

(The Elements: Book $\text{III}$: Proposition $13$)


Proof

Aiming for a contradiction, suppose it is possible for two circles to touch at more points than one.

Euclid-III-13.png

First, let the circle $ABDC$ touch the circle $EBFD$ internally at more than one point, that is, at $B$ and $D$.

Let $G$ be the center of the circle $ABDC$, and $H$ be the center of the circle $EBFD$.

(It is clear that in the diagram these centers are not actually at $G$ and $H$, and in fact $EBFD$ is obviously not a circle - it is the point of this proof to demonstrate that this would not be possible.)

From Line Joining Centers of Two Circles Touching Internally the straight line $GH$ will pass through both $B$ and $D$.

Since $G$ is the center of the circle $ABDC$, we have:

\(\ds BG\) \(=\) \(\ds GD\)
\(\ds \leadsto \ \ \) \(\ds BG\) \(>\) \(\ds HD\)
\(\ds \leadsto \ \ \) \(\ds BH\) \(\gg\) \(\ds HD\)

But since $H$ is the center of the circle $EBFD$, we have that $BH = HD$.

But we have just shown that $BH \gg HD$, which is impossible.

Therefore a circle does not touch another circle internally at more than one point.


Next suppose the circle $ACK$ touches the circle $ABDC$ at more points than one, that is, at $A$ and $C$, and join $AC$.

The two points $A$ and $C$ fall on the circumference of both circles $ABDC$ and $ACK$.

So it follows from Chord Lies Inside its Circle that $AC$ lies with both circles.

But from Book $\text{III}$ Definition $3$: Tangent Circles this line would fall inside $ABDC$ and outside $ACK$.

This reveals a contradiction.

Therefore a circle does not touch another circle externally at more than one point.

$\blacksquare$


Historical Note

This proof is Proposition $13$ of Book $\text{III}$ of Euclid's The Elements.


Sources