Two Circles have at most Two Points of Intersection

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

Geometric Proof
Suppose the opposite, and let circle $$ABC$$ cut the circle $$DEF$$ at more points than two, that is at $$B, G, F, H$$.


 * Euclid-III-10.png

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.

From Intersecting Circles Have Different Centers, this statement is contradictory.

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