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Re: Coordinates for circle

  • To: mathgroup at
  • Subject: [mg14585] Re: Coordinates for circle
  • From: Daniel Lichtblau <danl>
  • Date: Mon, 2 Nov 1998 01:50:58 -0500
  • Organization: Wolfram Research, Inc.
  • References: <71blgi$>
  • Sender: owner-wri-mathgroup at

MAvalosJr at wrote:
> Hi Everyone
> Maybe I can restate my problem of two days ago since I didn't get any
> replies. Given two coordinates on a circle, for example,{0,0},{2,2},
> and the radius of the
> circle, i.e.,2 Sqrt[5], to get a third set of coordinates. With this
> third set, using the standard Quadric form for a circle,i.e., a x^2 + a
> y^2 + b x + c y + d ==0, using the Conics from mathematica I can get
> the equation of the circle- which is what I'm looking for.   Thanks for
> whatever.
> Manuel

It may be simpler just to find the circle without grabbing a third
coordinate thereon. Also note the data does not give a unique circle.
This is geometrically obvious because if you obtain a center you can
reflect it across the segment that joins your pair of coordinates and
have another circle with the same radius that hits those coordinates.

To represent the circle in the form you desire, denote the center as
{cx,cy} and expand the usual equation.

In[4]:= circ = Expand[(x-cx)^2 + (y-cy)^2 - 20];

We have two coordinates that satisfy this equation, use then to find

In[5]:= soln = Solve[{cx^2 + cy^2==20, (2-cx)^2 + (2-cy)^2==20},
        {cx, cy}];

Plug the result into the expanded circle equation.

In[6]:= InputForm[circ /. soln]
Out[6]//InputForm= {4*x + x^2 - 8*y + y^2, -8*x + x^2 + 4*y + y^2}

Daniel Lichtblau
Wolfram Research

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