Dem.—If possible let them have a common centre at O. Join OA, and draw any other line OD, cutting the circles in C and D respectively. Then because O is the centre of the circle ABC, OA is equal to OC. Again, because O is the centre of the circle ABD, OA is equal to OD. Hence OC is equal to OD—a part equal to the whole—which is absurd. Therefore the circles are not concentric.
Exercises.
1. If two non-concentric circles intersect in one point, they must intersect in another point. For, let O, O′ be the centres, A the point of intersection; from A let fall the ⊥ AC on the line OO′. Produce AC to B, making BC = CA: then B is another point of intersection.
2. Two circles cannot have three points in common without wholly coinciding.
PROP. VI.—Theorem.
If one circle (ABC) touch another circle (ADE) internally in any point (A), it is not concentric with it.
Dem.—If possible let the circles be concentric, and let O be the centre of each. Join OA, and draw any other line OD, cutting the circles in the points B, D respectively. Then because O is the centre of each circle (hyp.), OB and OD are each equal to OA; therefore OB is equal to OD, which is impossible. Hence the circles cannot have the same centre.
PROP. VII.—Theorem.