|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.||I.10|
|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.||III.1,Cor|
|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.||III.5|
|Therefore a circle does not cut a circle at more than two points.|
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.
Next proposition: III.11