 | | To all my friends! —Dave Schuler
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| | | The way I show my gratitude when I receive informational help from people is by shamelessly exploiting their knowledge again, so here goes: If I know the lengths of two chords of a circle, both of which are smaller than the diameter and which are (...) (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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| | | | | Re: To all my friends! —Larry Pieniazek
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| | | | (...) I am just guessing, but don't 4 points uniquely (over) determine a circle, because any 3 noncollinear points actually uniquely determine a circle? If you can calculate the radius from any 3 non collinearpoints you can do it from this info as (...) (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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| | | | | | Re: To all my friends! —Rob Doucette
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| | | | (...) r = sqrt[(A^2/8D + D/2 - B^2/8D)^2 + B^2/4] where: A = chord 1 length B = chord 2 length D = perpindicular distance between A and B or you could search on cyclic quadrilateral and solve for r using that mess, simplifying for the quadrilateral (...) (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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| | | | | | Re: To all my friends! —Dave Schuler
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| | | | (...) Cool. Thank you to both of you--I can work with that formula. Dave! (23 years ago, 7-Nov-01, to lugnet.off-topic.geek)
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