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Re: How to get the solution area of inequation?

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  • Subject: [mg12071] Re: How to get the solution area of inequation?
  • From: Paul Abbott <>
  • Date: Sat, 25 Apr 1998 01:30:09 -0400
  • Organization: University of Western Australia
  • References: <6h70nf$>

8623806 wrote:

> I have three circles:
> C1: (0,0) with radius r
> C2: (a,0) with radius r/k
> C3: (a+b Cos[\[Theta]], b Sin[\[Theta]]) with radius r/k
> k is an integer and 0 <  a< r, 0 < [Theta] < Pi, also the center of C3
> is in the intersection of C1 and C2.
> If there is a possible general solution for C1 && C3 && !C2 ??

Are you after an expression that gives the set of all points in C1 and
C3 that are not in C2?  In general, this is a reasonably complicated
set and I'm not even sure how you think members of such a set might be
expressed?  For example, one way to describe the points inside the unit
circle is simply to test if x^2+y^2 < 1 -- and there is really no
simpler way to write (or compute) this.  In this simple case you could
parametrize the solutions as r < 1, 0 <= Theta < 2Pi -- but this is
about the same complexity.

Incidentally, the following code can be used to visualize the set for a
range of parameter values:

fig[r_, k_, a_, b_, t_] := 
   Show[Graphics[{Hue[0], Point[{0, 0}], Circle[{0, 0}, r], 
      Hue[0.3], Point[{a, 0}], Circle[{a, 0}, r/k], 
      Hue[0.7], Point[{a + b*Cos[t], b*Sin[t]}], 
      Circle[{a + b*Cos[t], b*Sin[t]}, r/k]}], 
    AspectRatio -> Automatic]; 


Paul Abbott                                   Phone: +61-8-9380-2734
Department of Physics                           Fax: +61-8-9380-1014
The University of Western Australia            Nedlands WA  6907  AUSTRALIA                   

            God IS a weakly left-handed dice player

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