This page uses JavaSketchpad.
You may drag any of the red points, and the construction will adjust accordingly.
Explanation of the construction:
The idea is to use inversion to convert two of the original three objects to parallel lines, and then
use construction 32. By considering the parallel line FZ and the circles AJ and BF, I get FZ and BF
throught the same point, which I use as the center of inversion. Note FZ inverts to itself. So
we find the center S of the circle tangent to FZ, HI, and RQ, and invert back to center X of
the desired circle.
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