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MathGroup Archive 2005

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Re: Circle equation problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg61538] Re: Circle equation problem
  • From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
  • Date: Sat, 22 Oct 2005 00:35:22 -0400 (EDT)
  • Organization: Uni Leipzig
  • References: <dja2c3$fig$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup 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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