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Topic:  Common Tangent Construction 
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Subject:  RE: Common Tangent Construction 
Author:  Mathman 
Date:  May 8 2006 
the radii are VERY close. I based that presumption on the fact that the
original solution would have to have a point FAR from the circles. The failure
lies in the fact that although not far form the circles, the second solution
requires a point of itnersection of two circles that becomes too small to
distinguish accurately.
Your solution is delightful, but it also fails [in inate accuracy] as the
difference circle becomes too tiny to distinguish. Great in theory though, as
my own seemed to be.
I'll describe my approach, which also works well for circles of reasonable
difference in diameter. Assume two circles, centered on a horizontal line,
larger to the left so we are looking at the same thing:
Method 1.
Draw a radius of one circle [e.g. up to the right.]
Draw one in the second, parallel to the first. Extend the line joining points
of intersection of radii with the circles to meet the [horizontal] line joining
their centers at P.
Use that point with one of the far left points of an original circle diameter on
the horizontal line to draw a semicircle [full for both tangents.] It will
intersect the first circle at the required point of tangency. Join the far
point, P, to that for the common tangent.
Method 2. is actually more involved, and I was wrong in thinking it would
alleviate the problem in the above solution for P being too far to consider when
the two circles were very close in diameter. I'll go over it a bit more
thoroughly before I send that one here, just as an alternative approach. I
liked it because it involved a little algebra and the concept of mean
proportional within the semicircle.
Besides, the lawn took most of the day, finishing building and staining an oak
endtable the rest, and it's now late here. So, it will wait. :) Dentist
tomorrow, :(
David.
 
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