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

For, if possible, let the circle *ABC* cut the circle *DEF* at more points than two, namely *B, G, F,* and *H.*

Join *BH* and *BG,* and bisect them at the points *K* and *L.* Draw *KC* and *LM* from *K* and *L* at right angles to *BH* and *BG,* and carry them through to the points *A* and *E.*

Then, since in the circle *ABC* a straight line *AC* cuts a straight line *BH* into two equal parts and at right angles, the center of the circle *ABC* lies on *AC.* Again, since in the same circle *ABC* a straight line *NO* cuts a straight line *BG* into two equal parts and at right angles, the center of the circle *ABC* lies on *NO.*>

But it was also proved to lie on *AC,* and the straight lines *AC* and *NO* meet at no point except at *P,* therefore the point *P* is the center of the circle *ABC.*

Similarly we can prove that *P* is also the center of the circle *DEF,* therefore the two circles *ABC* and *DEF* which cut one another have the same center *P,* which is impossible.

Therefore *a circle does not cut a circle at more than two points.*

Q.E.D.

The proof actually shows that the two circles cannot *meet* in more than two points, where “meet” could be either cut or touch.

Heath remarks that the lines bisecting *BG* and *BH* have not been shown to meet. In fact, they have, since the center of the circle *ABC* has been shown to be on both.

This proposition is used in III.24.