Two Circles Have At Most Two Points of Intersection

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Theorem

A circle does not cut another circle at more points than two.

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. For the sake of demonstrating a contradiction, we suppose that 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.

Hence such a pair of circles, i.e. those that intersect at more points than two, can not exist.

$\blacksquare$

This is Proposition 10 of Book III of Euclid's The Elements.