On Tue, Mar 23, 2010 at 07:15:30PM -0400, Przemek Klosowski wrote: > I haven't cross-checked this, but inputting this into Maxima: > > solve([(x1-h)^2 + (y1-k)^2 = r^2, (x2-h)^2 + (y2-k)^2 = r^2, (x3-h)^2 > + (y3-k)^2 = r^2], [h, k, r]); > > results in: > > [h=-((y2-y1)*y3^2+(-y2^2+y1^2-x2^2+x1^2)*y3+y1*y2^2+(-y1^2+x3^2-x1^2)*y2+(x2^2-x3^2)*y1)/((2*x2-2*x1)*y3+(2*x1-2*x3)*y2+(2*x3-2*x2)*y1), > k=((x2-x1)*y3^2+(x1-x3)*y2^2+(x3-x2)*y1^2+(x2-x1)*x3^2+(x1^2-x2^2)*x3+x1*x2^2-x1^2*x2)/((2*x2-2*x1)*y3+(2*x1-2*x3)*y2+(2*x3-2*x2)*y1),
This looks pretty reasonable for the hole center (although I didn't check or try it), but the solution for r (which I clipped) is absurd. You can find r using the distance formula given hole center h,k and any one of the original points. In this case a simultaneous solution is not needed, and our knowledge of the problem's geometry can lead us to a much simpler answer. ------------------------------------------------------------------------------ Download Intel® Parallel Studio Eval Try the new software tools for yourself. Speed compiling, find bugs proactively, and fine-tune applications for parallel performance. See why Intel Parallel Studio got high marks during beta. http://p.sf.net/sfu/intel-sw-dev _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
