In article ,
James Waldby wrote:

On Sun, 08 May 2011 10:28:21 -0700, Tim Wescott wrote:
On 05/03/2011 12:59 PM, Tim Wescott wrote:
Consider two circles, of arbitrary diameter, and a point, all on a
plane.

I want to inscribe an arc that is tangent to both circles, and which
passes through the point.

Anyone know a way to construct the arc? I'm not snickering in the
background here as I pose puzzles -- this is a drafting problem that
I'm running into quite a lot lately.

Well, I still haven't figured out a solution for the general problem
(and if there's a way to get QCad to do it I haven't found that either).
But after days of fumbling with math and compasses I found a solution
for two _equal_ sized circles and a point. [...]
Inscribe your two circles, call their centers "A" and "B". Define your
point on the arc as "C". Draw line AB between the circle centers.

Now make line CD that's perpendicular to line AB (and passes through C).
Find the point on line CD that's one circle radius inward of point C
-- call this point E.

Now make an arc between points A, B, and E (QCad does have an "arc from
three points" function).

Now make an arc parallel to arc AEB, and larger by one circle radius.
It'll be tangent to the two circles, and pass through point E.

Presumably you meant "pass through point C" at the end there. The larger
arc will go through C when point D bisects line AB, but not otherwise.

Note, your problem is the CCP form of Apollonius' Problem, as noted in (eg)
http://en.wikipedia.org/wiki/Special...blem#Type_7:_T
wo_circles.2C_one_point_.28CCP.2C_generally_4_solu tions.29
(About 80% thru
http://en.wikipedia.org/wiki/Special_cases_of_Apollonius'_problem)

Cute. It figures that the ancients solved the problem, with only
straightedge and compass. It was a competitive sport back then.

Joe Gwinn