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!!! :-)
|