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