Problem of Apollonius
Many special cases of Apollonius' problem involve finding a circle that is tangent to one or more lines. The simplest of these is to construct circles that are tangent to three given lines (the LLL problem). To solve this problem, the center of any such circle must lie on an angle bisector of any pair of the lines; there are two angle-bisecting lines for every intersection of two lines. The intersections of these angle bisectors give the centers of solution circles. There are four such circles in general, the inscribed circle of the triangle formed by the intersection of the three lines, and the three exscribed circles.
A general Apollonius problem can be transformed into the simpler problem of circle tangent to one circle and two parallel lines (itself a special case of the LLC special case). To accomplish this, it suffices to scale two of the three given circles until they just touch, i.e., are tangent. An inversion in their tangent point with respect to a circle of appropriate radius transforms the two touching given circles into two parallel lines, and the third given circle into another circle. Thus, the solutions may be found by sliding a circle of constant radius between two parallel lines until it contacts the transformed third circle. Re-inversion produces the corresponding solutions to the original problem.
Read more about this topic: Tangent Lines To Circles
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