|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.||I.Def.15|
|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.|
This proposition is used in III.10 which states that circles cannot intersect at more than two points.
Next proposition: III.6