Re: How to get the solution area of inequation?
- 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
____________________________________________________________________