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Re: Coordinates for circle
*To*: mathgroup at smc.vnet.net
*Subject*: [mg14603] Re: [mg14579] Coordinates for circle
*From*: BobHanlon at aol.com
*Date*: Mon, 2 Nov 1998 01:51:10 -0500
*Sender*: owner-wri-mathgroup at wolfram.com
In a message dated 10/30/98 6:40:00 AM, MAvalosJr at AOL.com writes:
>Maybe I can restate my problem of two days ago since I didn't get any
>replies. Given two coordinates on a circle, for example,{0,0},{2,2},
>and the radius of the
>circle, i.e.,2 Sqrt[5], to get a third set of coordinates. With this
>third set, using the standard Quadric form for a circle,i.e., a x^2 + a
>y^2 + b x + c y + d ==0, using the Conics from mathematica I can get
>the equation of the circle- which is what I'm looking for. Thanks for
>whatever.
>
Manuel,
Needs["Graphics`ImplicitPlot`"]
a can be set to any arbitrary value, say a = 1. Also, if the origin of
the circle is at {e, f}, then the point on the circle opposite the
point {0, 0} is the point {2e, 2f}
Solve[Flatten[{a == 1,
{a x^2 + a y^2 + b x + c y + d == 0,
(x-e)^2 + (y-f)^2 == (2 Sqrt[5])^2} /.
{{x -> 0, y -> 0}, {x -> 2, y -> 2}, {x -> 2e, y -> 2f}}}],
{a, b, c, d, e, f}]
{{d -> 0, b -> -8, c -> 4, a -> 1, f -> -2, e -> 4},
{d -> 0, b -> 4, c -> -8, a -> 1, f -> 4, e -> -2}}
a x^2 + a y^2 + b x + c y + d == 0 /. %
{x^2 - 8*x + y^2 + 4*y == 0, x^2 + 4*x + y^2 - 8*y == 0}
ImplicitPlot[%, {x, -10, 10}];
Bob Hanlon
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