Given a circle C, a line L that doesn't intersect C, and a point P that is not on L and not inside C,...

I first saw that one back in high school, many (too many) years ago.
The solution was a complicated ruler-and-compass construction.
However, it is easy to understand such a solution if you use the idea
of inversion, which I didn't know about way back then.

First, P and C must both be on the same side of L, for otherwise there
is obviously no solution.  Let X denote the unknown circle we have to
find.  Invert with centre P and any convenient radius.  Let L', C' and
X' denote the inverses of L, C and X respectively.

Then L' is a circle passing through P, and it is easy to construct it.
C' is another circle, which is also easily constructed.  The unknown X'
will be a straight line that is tangent to both L' and C'.  After you
have drawn L' and C', it is easy to find all their common tangent
lines, and X' can be any one of them.  Then invert X' to get X.  This
shows that there should be four possible answers.

(Parenthetical remark -- we are given that L does not intersect C, so
L' does not meet C', and these two circles will have four common
tangent lines.  However, this approach will also work even if L meets
C, as long as P is outside C.  In this more general case, there will be
fewer than four answers.)

There are many steps involved, so the full detailed construction will
be lengthy.  However, this approach makes it easy to understand that
lengthy step-by-step construction.  One simplification results from
choosing a good value for "any convenient radius".  Let c be the centre
of C, and let D be a point on C such that the line PD is perpendicular
to the line Dc.  (You can construct D by drawing a circle with diameter
Pc.)  If you use PD as the radius of inversion, then you get the
simplification that C' = C, so the steps used to construct C' can be
omitted.

-- Barry Wolk <wolk@ccm.umanitoba.ca>

[ Editor: the problem was posted by Jonathan Haas
<positron@Starbase.NeoSoft.COM> from a Microsoft internal puzzles
group. ]