Re: Transformstion to canonical form

*To*: mathgroup at smc.vnet.net*Subject*: [mg65504] Re: Transformstion to canonical form*From*: dh <dh at metrohm.ch>*Date*: Wed, 5 Apr 2006 06:56:06 -0400 (EDT)*References*: <e0j48b$g7l$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

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: w=u^2+v^2 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 Daniel 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 >