Two Circles have at most Two Points of Intersection
Theorem
In the words of Euclid:
(The Elements: Book $\text{III}$: Proposition $10$)
Proof
Suppose the opposite, and let circle $ABC$ cut the circle $DEF$ at more points than two, that is at $B, G, F, H$.
Clearly, in the diagram, circle $DEF$ is not actually a circle.
Aiming for a contradiction, suppose it is.
Join $BH$ and $BG$, and bisect them at $K$ and $L$.
From $K$ and $L$, draw perpendiculars to $BH$ and $BG$, through to $A$ and $E$.
Since $AC$ cuts $BH$ into two equal parts at right angles, from the porism to Finding Center of Circle, the center of circle $ABC$ is on $AC$.
For the same reason, the center of circle $ABC$ is also on $NO$.
But as the only point on both $AC$ and $NO$ is $P$, it follows that $P$ is the center of circle $ABC$.
Similarly, we can show that $P$ is also the center of circle $DEF$.
So we have two circles that intersect which have the same center.
From Intersecting Circles have Different Centers, this statement is contradictory.
Hence such a pair of circles, that is, those that intersect at more points than two, can not exist.
$\blacksquare$
Historical Note
This proof is Proposition $10$ 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