A Planar Geometry Problem

First: in general, it is not solvable. There are points which cannot lie on a circle tangent to two others. E.g., the point cannot be inside either circle. And there are 4 other, "small", areas that are difficult to describe, but are easily shown on a graphic (later).

Second, and more importantly, the radius of the solution circle can approach infinity. So, in practical terms, it becomes unsolvable at some point, depending upon whether you are using CAD or a drawing board .

Now, there are three cases for the solution: where the given circles are externally or internally tangent to the solution circle, or one of each. Do your problems always fall into one of these classes? Also, if the given circles can overlap, it may be a different set of solution classes (I haven't thought that through). Do your circles ever overlap?

I love plane geometry, Bob

My approch:

One assues three circles. C1, C2 and Carc

which can be defined by x1,y1, R1 x2,y2,R2 and xa,ya,Ra Carc is the curve you are looking for

a tanjent unit vector on each circule is defiend as ((2(x-xn)x) i

+(2(y-Yn)y )j)/(Sqrt( ((2(x-xn))^2 +(2(y-Yn)y)^2) where i and j are unit vectors in the x and y direction

A tanjent condition is acheved when the C1, Carc share the same point and tanjent vectors are colinear. Plug the equations into Mathmaica and pray

Dear Tim,

Maybe I do not understand the problem. I seem to find cases for which the is no such arc.

Start with one arbitrary point (0,0) and one circle of an arbitrary position of the center (2,0) and radius R1 = 1. In this case the "solution" is a circle centered at (0,0) with a radius of 1.

TO do this by construction one draws a line between the point and the center of the circle. Set the spread of one's compass to the length of the line from the point to the intersection of the line and the circle. Swing your arc.

The desired arc is specified by the po> >> Consider two circles, of arbitrary diameter, and a point, all on a plane. >>

Don't know how to do it with compass and ruler, but I use Rhino3d and it takes about 5 seconds if you have the 2 circles and point already defined. If the point and circles are reasonably positioned there are 4 possible solutions. You can influence which solution you get by where you pick on the 2 circles.

you can download a trial version for free from Rhino3d.com

-jim

Is that not a three point circle - but don't complete the circle.

The three points are The point, and the two tangent points.

search for topics containing "equations of a circle"

Mart> Consider two circles, of arbitrary diameter, and a point, all on a plane. >

No. The three point circle won't in many cases, be tangent to the other circles at those two points.

jk

The problem is finding the tangent points.

In article , Tim Wescott writes

Just tried it using an ancient version of EasyCad. I just drew 3 circles and used the "draw circle tangent to 3 circles" instruction, and then trimmed the new circle to get the required arc.

The third circle can be made very very small so that in practical terms it approximates to a point.

In autocad you can draw a circle tan, tan ,tan to the two circles and the point

In Solidworks you can draw an arc through the point then constrain it to be tangent to each of the two circles

Best regards, Spehro Pefhany

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 here:

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.

I assume that the arc is circular.

It's either easy or it's impossible. Most possible locations of Point X (the center of the circular arc that's tangent to the two original circles and passes through Point C) do not have a solution.

Turning this around, using algebra, you can derive the function giving the space of locations for which a solution exists.

Does the arc need to be circular, or will a Bezier curve do?

Joe Gwinn

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)

(About 80% thru

It's for a CAD program that handles ellipses badly, and Bezier's even worse.

[snip description of not-quite-a-solution]