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MathGroup Archive 2006

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Re: Transformstion to canonical form

  • To: mathgroup at
  • Subject: [mg65504] Re: Transformstion to canonical form
  • From: dh <dh at>
  • Date: Wed, 5 Apr 2006 06:56:06 -0400 (EDT)
  • References: <e0j48b$g7l$>
  • Sender: owner-wri-mathgroup at

Hi Wolfgang,
(u,v,w) describe an arbitrary circle, whereas (G,b) describe a circle 
through the origin. Therefore, an additional equation must exist, namely:

To make life a bit easier (without loss of generality), I would advice 
to use a symmetrical arrangement, z.B.
x1 = {-b, 0}; x2 = {b, 0}; x0 = {x, y};
This makes u=0, what leaves only v and w
With this we get (I square to get rid of square roots

eq1 = ((x0-x1).(x0-x2))^2==G^2 (x1-x0).(x1-x0) (x2-x0).(x2-x0);
eq2 = x^2 + (y - v)^2 == w^2;

Reduce will then give quit a lot of possible solutions:

Reduce[{eq1, eq2}, {v, w}, {x, y}]

on of which e.g.:

{w -> -(v/G), v -> (b*G)/Sqrt[1 - G^2]}

where G^2-1=!=0 and G=!= 0


Dr. Wolfgang Hintze wrote:
> The following rather simple problem led my to some genral questions of 
> how to proceed in symbolic transformations which I would do on paper 
> within Mathematica. I'm sure you can give useful hints which I apreciate 
> in advance.
> Consider a slight generalization of the Thales circle: In a triangle 
> ABC, let g be the fixed angle in vertex C. (In the Thales case we have g 
> = Pi/2).
> What is the equation for the coordinates (x,y) of the point C?
> The immediate statement is derived from the scalar product of the two 
> vectors (C-A) and (C-B):
> (1) 
> (C-A).(C-B)==  Sqrt[(C-A).C-A)] Sqrt[(C-B).(C-A)]
> where
> (2) 
> G = Cos[g]
> Putting
> A={0,0}; B={b,0};C={x,y} we obtain the equation:
> eq1 = x^2 - b*x + y^2 == G*Sqrt[x^2 + y^2]*Sqrt[(x - b)^2 + y^2]
> Now: what is a useful procedure in Mathematica to transform eq1 into an 
> equation of the form
> eq2 = (x - u)^2 + (y - v)^2 == w
> and to find the parameters u, v, and w in terms of b and G?
> Regards,
> Wolfgang

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