Clifford's circle chain theorem with n=7 circles

Following Penrose' notation, I omit letters "c" for circles and "P" for points, to avoid cluttering. Here we see seven (n=7) blue circles c1, c2, c3, c4, c5, c6, c7 through a point O (taken to infinity) intersecting in P12, P13, P14, P15, P16, P17, P23, ..., P67 which again define unique yellow circles like as c123, c124, ..., c567 (not all drawn, to avoid cluttering). The construction of these yellow circles corresponds to applying the case n=3 successively to three circles each, i.e. to c1,c2,c3 to give c123, etc. According to the case with n=4 circles through O taking four yellow circles at a time yields the small dark red points P1234, P1235, ..., P4567. The case n=5 shows that always five of these points at a time, e.g. P1234, P1235, P1245, P1345, P2345 lie on red circles c12345, etc. According to the case with n=6 allways six such red circles, e.g. c12345, c12346, c12356, c12456, c13456, c23456 will coincide in points like P123456. The surprise in the case of n=7 below is that the total of seven such points P123456, P134567, P124567, P123567, P123467, P123457, P123456 lie on one unique (violet) circle c1234567.

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

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