| It may help to realize that the minimum distance is the diameter of the
smallest sphere that is tangent to both circles. This is a generalization
of the theorem that the minimum distance between two smooth closed curves
in the plane is normal, i.e. perpindicular to the tangents, to both curves.
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| One way to approach the problem is by considering a generic point on one
circle, and minimize the distance from this point to a point on the other
circle (there are either 1 or 2 minima, or the entire other circle is equi-
distant from the point). Then minimmize the distance with respect to the
generic point on the first circle.
It would help if the problem could be transformed so that one of the circles
were perpendicular to the line joing their centers, and which kept the
distance formula relatively simple. Ideas?
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| I played with this one a bit and suspect that there may not be a simple
formula. I tried representing the solution in algebraic terms, as follows:
With no loss in generality, place one circle, say the larger, at the
origin of the XY plane with radiius 1. Then let the other lie in a plane
spanned by a pair of orthogonal unit vectors, calling that the VH (vertical
horizontal) plane, with center at (x0,y0,z0), and radiius r.
If the unit vectors Uh and Uv have the XYZ components (uhx,uhy,uhz) and
(uvx,uvy,uvz) and we demand the distance vector from the XY circle to the
VH circle be normal to the tangent vectors on the two circles this pair of
equations easily results:
x*y0 - y*x0 + x*h*uhy + x*v*uvy - y*h*uhx - y*v*uvx = 0
h*(x0*uvx + y0*uvy + z0*uvz) - v*(x0*uhx + y0*uhy + z0*uvz)
- x*h*uvx + x*v*uhx - y*h*uvy + y*v*uhy = 0
and, x^2 + y^2 = 1, h^2 + v^2 = r^2.
So you have a bilinear set of equations in (x,y,v,h), but it doesn't look
like it will simplify much. Parameterizing the (x,y) and (v,h) in terms
of trig functions will reduce the number of variables, but the resulting
equations don't look any simpler, though there are some intersting symmetries.
Well, not very illuminating so far...
- Jim
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