Subject:
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Re: Here's looking at Euclid
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Newsgroups:
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lugnet.off-topic.geek
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Date:
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Wed, 2 Aug 2000 00:32:25 GMT
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Viewed:
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231 times
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In lugnet.off-topic.geek, Dave Schuler writes:
> Do I understand you correctly that with three points we could calculate the
> circle? Would you mind giving a brief (if possible) description of the
> process in that case?
The process depends on whether you want a numerical process or a construction
process. I'm going to assume that you want a numerical process and that you
know the coordinates of the three points. If you want a construction process
to be able to draw such a circle with a straight-edge and compass, let me know.
Let's call the three points A, B, and C. First check to make sure they aren't
colinear - that they don't all lie on the same line. If they do, then you
can't make a circle (unless you want to allow circles with infinite radius).
You should already have coordinates for the three points. Let's use (x1, y1)
for the coordinates of point A, (x2, y2) for point B, and (x3, y3) for point
C. Then all you have to do is use three instances of the distance formula to
get a system of two equations with two unknowns, and then solve the system of
equations. All of that gets very messy, so I'll just give the results.
To make the math (a little) less complicated, I'm going to use some
simplifying expressions:
P=(x1-x2)/(y2-y1)
Q=(x2^2-x1^2+y2^2-y1^2)/(2*(y2-y1))
R=(x1-x3)/(y3-y1)
S=(x3^2-x1^2+y3^2-y1^2)/(2*(y3-y1))
Then the x-coordinate (let's call it X) of the center of the circle is (S-Q)/
(P-R) and the y-coordinate (let's call it Y) is (P*(S-Q)/(P-R))+Q. And the
radius (let's call it R) of the circle is sqrt((x1-X)^2+(y1-Y)^2).
Then the equation of the circle is (x-X)^2+(y-Y)^2=R^2.
I hope this is what you were looking for and that's it's understandable. I
believe I did the equation manipulation correctly, but would love confirmation
of this from some other math geek.
If I can make this any clearer, let me know.
John Gramley
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Message has 1 Reply:
 | | Re: Here's looking at Euclid
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| In lugnet.off-topic.geek, John Gramley writes: **snip of some rather helpful stuff** How about this: Suppose the two points are vertices of an inscribed octagon whose sides are each of length X. Would that help? Dave! (24 years ago, 2-Aug-00, to lugnet.off-topic.geek)
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Message is in Reply To:
| | Re: Here's looking at Euclid
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| (...) Yeah, I was afraid of that. Believe it or not, almost immediately after I posted I was sitting at a circular table with a can of Coke, and I realized that the same 1 1/2 inch that defines the diameter of the can only describes a tiny portion (...) (24 years ago, 1-Aug-00, to lugnet.off-topic.geek)
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