Proof: By Euclid
(related to Proposition: 3.05: Intersecting Circles have Different Centers)
 For, if possible, let $E$ be (the common center), and let $EC$ have been joined, and let $EFG$ have been drawn through (the two circles), at random.
 And since point $E$ is the center of the circle $ABC$, $EC$ is equal to $EF$.
 Again, since point $E$ is the center of the circle $CDG$, $EC$ is equal to $EG$.
 But $EC$ was also shown (to be) equal to $EF$.
 Thus, $EF$ is also equal to $EG$, the lesser to the greater.
 The very thing is impossible.
 Thus, point $E$ is not the (common) center of the circles $ABC$ and $CDG$.
 Thus, if two circles cut one another then they will not have the same center.
 (Which is) the very thing it was required to show.^{1}
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes