Let a = (O, OB), b = (C, CB) and c = (D, DE) be three circles. (a) and (b) are tangent at B. All three have fixed radius and B moves on (a), while (b) remains tangent to (a) at B all the time. For each position of B the bisectors of angle AOB are drawn and their intersections F, G with the radical axis of the circles (b) and (c) are defined. Show that F and G move on two lines orthogonal to the common diameter AD of circles (a) and (c).
The proof is trivial, using coordinates with origin at O. The following geometric proof is due to Antreas Varverakis. Consider the circles d = (C',C'H), e = (C'',C''A), which are symmetric of (b) with respect to the lines OG and OF respectively. Obviously G and F are radical centers of the tripples of circles: ((b),(d),(c)) and ((b),(e),(c)) respectively. Since, in each tripple, the two last circles are fixed, the lines [LG] and [KF] are also fixed.
The line [FG] envelopes an ellipse with one focus at O. Look at ThreeCirclesProblem1.html if you are interested in.
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