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MathGroup Archive 1998

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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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