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Re: Circle equation problem
 To: mathgroup at smc.vnet.net
 Subject: [mg61538] Re: Circle equation problem
 From: "JensPeer Kuska" <kuska at informatik.unileipzig.de>
 Date: Sat, 22 Oct 2005 00:35:22 0400 (EDT)
 Organization: Uni Leipzig
 References: <dja2c3$fig$1@smc.vnet.net>
 Sender: ownerwrimathgroup at wolfram.com
expr = Abs[p]^2 == Abs[(G  X + I*Y)/(1 + G*(X +
I*Y))]^2;
expr1 = ComplexExpand[expr, TargetFunctions >
{Re, Im}] // FullSimplify;
expr2 = (Expand /@ (expr1 /. a_ == b_/c_ :> a*c ==
b)) /.
a_ == b_ :> a  b == 0
and you see that at least X must be scaled by
x>X/(p*G)
Regards
Jens
"Daniele Lupo" <danwolf80_no_spam_ at libero.it>
schrieb im Newsbeitrag
news:dja2c3$fig$1 at smc.vnet.net...
 Hi to all.

 I've this equation:

 Abs[p]^2 == Abs[(G  X + I*Y)/(1 + G*(X +
I*Y))]^2

 It represent a circle in the complex plane; X
and Y are coordinates in the
 complex plane (point is X + I Y), while G, p are
complex parameters.

 I want to write it in this standard form for a
circle in the cartesian
 plane:

 (X  Xcenter)^2 + (Y  YCenter)^2  ray^2 == 0

 In this way, the center in the complex plane is
XCenter + I YCenter,

 With some effort I think to be able to solve it
by hand, but I want to know
 if there's a (easy?) way to find center and ray
of the circle with
 Mathematica. How can I find Xcenter, YCenter and
ray from the original
 equation?

 Thanks for your answers.

 Daniele Lupo

 PS: Thank you for all your replies in my
previous posts!!! :)

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