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The statement is true. We prove it in three small bits. For an alternative proof, see posts by Tristan and Joel.
1. If suffices to show that f maps collinear points to collinear points.
Proof: If l is a line then f(l) is at least a sub set of a certain line l'. If there were a point P' on l' such that P:=f-1(P') is outside l, then pick points A and B on l and a circle through A, B and P (There is always a circle through three non-collinear points). But then f(A), f(B) and f(P)=P' lie on a circle, which contradicts that P' is on l'. Therefore f(l)=l'
2. f-1 maps collinear points to collinear points.
Proof: Otherwise there are non-collinear points A, B, C such that A'=f(A), B'=f(B) and C'=f(C) are collinear. But A', B' and C' must lie on a circle, because A, B and C do.
3. f maps collinear points to collinear points.
Proof:
Consider three collinear points A, B and P, where P is in between A and B. Let C be the circle through A and P with diameter AP and D the circle through B and P with diameter BP. Denote the images under f by A', B', P', C' and D'. The line through A' and P' intersects D' in a point S'. By statement 2 we know that S:=f-1(S') is on the line through A and P. S is also on the circle D. Since S is unequal P, we have S=B and therefore S'=B'. Hence B' is on the same line as A' and P'.
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