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. It's so easy that it made me
feel stupid for a good hour. Kind of like spending a day
troubleshooting a TV for every problem known to man, then finding out
that the reason the power supply appeared to be shot was because it was
unplugged.

So, in spite of my embarrassment, I share the answer he

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.

I'm still looking for a way to do this with circles of different
diameter -- hopefully it's almost as easy.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html