Clifford's circle chain theorem with n=6 circles

Following Penrose' notation, I omit letters "c" for circles and "P" for points, to avoid cluttering. Here we see six (n=6) blue circles c1, c2, c3, c4, c5, c6 through O intersecting in P12, P13, P14, P15, P23, P24, P25, P34, P35, P45, ..., P56 which again define unique yellow circles like as c123, c125, c145, c234, c345, ..., c456. The construction of these yellow circles corresponds to applying the case n=3 successively to three (blue) circles each, i.e. to c1,c2,c3 to give c123, etc. According to the case with n=4 circles through O, successively taking four yellow circles at a time yields the gold colored points P1234, P1235, P1245, P1345, P2345, ..., P3456. The case n=5 shows that always five of these golden points at a time, e.g. P1234, P1235, P1245, P1345, P2345 lie on red circles c12345, etc. Alltogether there will be six such red circles c12345, c12346, c12356, c12456, c13456, c23456. The surprising fact of the n=6 case is, that these six red circles all coincide in the (fat violet) point P123456.

The bright red points can be moved interactively with the mouse pointer.

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The six blue straight lines are now six circles with infinite radius intersecting at infinity O.

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Soli Deo Gloria. Created with Cinderella by Eckhard Hitzer (Fukui).