If two circles cut one another, then they do not have the same center.

Let the circles *ABC* and *CDG* cut one another at the points *B* and *C.*

I say that they do not have the same center.

For, if possible, let it be *E.* Join *EC,* and draw *EFG* through at random.

Then, since the point *E* is the center of the circle *ABC, EC* equals *EF.* Again, since the point *E* is the center of the circle *CDG, EC* equals *EG.*

But *EC* was proved equal to *EF* also, therefore *EF* also equals *EG,* the less equals the greater which is impossible.

Therefore the point *E* is not the center of the circles *ABC* and *CDG.*

Therefore *if two circles cut one another, then they do not have the same center.*

Q.E.D.

This proposition is used in III.10 which states that circles cannot intersect at more than two points.