*To*: mathgroup@smc.vnet.net*Subject*: [mg12071] Re: How to get the solution area of inequation?*From*: Paul Abbott <paul@physics.uwa.edu.au>*Date*: Sat, 25 Apr 1998 01:30:09 -0400*Organization*: University of Western Australia*References*: <6h70nf$c@smc.vnet.net>

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]; Cheers, Paul ____________________________________________________________________ Paul Abbott Phone: +61-8-9380-2734 Department of Physics Fax: +61-8-9380-1014 The University of Western Australia Nedlands WA 6907 mailto:paul@physics.uwa.edu.au AUSTRALIA http://www.pd.uwa.edu.au/~paul God IS a weakly left-handed dice player ____________________________________________________________________