Geometry problem. How can I inscribe a circle inside a figure consisting of three intersecting arcs?

Question

The arcs will either be all convex inward or all concave inward and the inscribed circle must be exactly touching all three arcs. I can get close using AutoCAD and the "trial and error" method, but it seems that there would be a formula or a drafting procedure that would make it simple, quick and accurate every time. If there are any math geniuses out there, consider this a challenge. I have not been able to solve this for a few days now.

Answer 1

The radius of two tangent circles are both perpendicular to the (common) tangent at the point of tangency, then the point of tangency and both centers lie on the same straight

line.

The problem is to find the coordinates of the center and the radius of the incribed circle. We have three unknown numbers, then we have to write down three equations.

The distance from the center of the inscribed circle to the three points of tangency is the same, and is the radius. Using the fact that the centers and the point of tangency are

colinear, the radii and the distance between centers must satisfy that:

| r1 ± √[(x1-x4)²+(y1-y4)²] | = r4

Where x4, y4, and r4 are the coordinates of the center and the radius of the inscribed circle (the unknowns), and x1, y1, r1 the center and radius of the first arc.

± because I don't know which center is closer to the point of tangency, and the absolute value because the tangency can occur "from the other side" if the arc is convex.

The other two equations are, obviously,

| r2 ± √[(x2-x4)²+(y2-y4)²] | = r4
| r3 ± √[(x1-x4)²+(y3-y4)²] | = r4

With a little algebra,

(x1-x4)² + (y1-y4)² = (r1 ± r4)² = (r1 + s1 r4)²
(x2-x4)² + (y2-y4)² = (r2 ± r4)² = (r2 + s2 r4)²
(x3-x4)² + (y3-y4)² = (r3 ± r4)² = (r3 + s3 r4)²

where s1, s2 and s3 can be +1 or -1.

Some algebra...

x1² - x2² - 2 (x1-x2) x4 + y1² - y2² - 2(y1-y2) y4 = r1² - r2² + 2(s1 r1 - s2 r2) r4
x1² - x3² - 2 (x1-x3) x4 + y1² - y3² - 2(y1-y3) y4 = r1² - r3² + 2(s1 r1 - s3 r3) r4

r4 = [x1² - x2² - 2 (x1-x2) x4 + y1² - y2² - 2(y1-y2) y4 - r1² + r2² ] / [2(s1 r1 - s2 r2)] =
= [x1² - x3² - 2 (x1-x3) x4 + y1² - y3² - 2(y1-y3) y4 - r1² + r3² ] / [2(s1 r1 - s3 r3)]


[(y1-y3) / (s1 r1 - s3 r3) - (y1-y2) / (s1 r1 - s2 r2)] y4 =
= [(x1-x2) / (s1 r1 - s2 r2) - (x1-x3) / (s1 r1 - s3 r3)] x4 +
+ [x1² - x3² + y1² - y3² - r1² + r3² ] / [2(s1 r1 - s3 r3)] -
- [x1² - x2² + y1² - y2² - r1² + r2² ] / [2(s1 r1 - s2 r2)]

Or

A y4 = B x4 + C,

where

A = [(y1-y3) / (s1 r1 - s3 r3) - (y1-y2) / (s1 r1 - s2 r2)]
B = [(x1-x2) / (s1 r1 - s2 r2) - (x1-x3) / (s1 r1 - s3 r3)]
and
C = [x1² - x3² + y1² - y3² - r1² + r3² ] / [2(s1 r1 - s3 r3)] -
- [x1² - x2² + y1² - y2² - r1² + r2² ] / [2(s1 r1 - s2 r2)]

y4 = (B/A) x4 + (C/A) = b x4 + c


r4 = [x1² - x2² - 2 (x1-x2) x4 + y1² - y2² - 2(y1-y2) (b x4 + c) - r1² + r2² ] / [2(s1 r1 - s2 r2)] =
= - [(y1-y2) b / (s1 r1 - s2 r2) + (x1-x2) / (s1 r1 - s2 r2)] x4 +
+ [x1² - x2² + y1² - y2² - 2(y1-y2) c - r1² + r2² ] / [2(s1 r1 - s2 r2)] =
= d x4 + e

where

d = - (y1-y2) b / (s1 r1 - s2 r2) - (x1-x2) / (s1 r1 - s2 r2)
and
e = [x1² - x2² + y1² - y2² - 2(y1-y2) c - r1² + r2² ] / [2(s1 r1 - s2 r2)]

and finally

x1² - 2 x1 x4 + x4² + y1² - 2 y1 (b x4+c) + (b x4+c)² = r1² + 2 s1 r1 (d x4+e) + (d x4+e)²

(1+ b² - d²) x4² + 2(b c - b y1 - x1 - d e - d s1 r1) x4 + (x1² + y1² - 2 y1 c + c² - r1² - 2 s1 r1 e - e²) = 0

or

p x4² + q x4 + r = 0,

where

p = (1+ b² - d²)
q = 2(b c - b y1 - x1 - d e - d s1 r1)
r = (x1² + y1² - 2 y1 c + c² - r1² - 2 s1 r1 e - e²)

By solving this quadratic equation we obtain x4, and then y4 and r4 from the previous equations.

Example:

x1=0, y1=3, r1=2
x2=2, y2=3/2, r2=5/2
x3=3/2, y3=2, r3=5/2

Solutions (excluding those with r4<0)

(x4,y4,r4)

1: (1.86591272426446 4.23346883607979 0.23675561118153)
2: (2.87178147865492 2.87178147865492 0.87464238651721)
3: (1.22289597024769 4.18617976560374 0.29632837953560)
4: (12.00000000000000 12.00000000000000 17.00000000000000)
5: ( -0.59018210164693 0.93044475985413 0.15206268615011)
6: ( 0.43256634743203 0.43256634743203 0.60361848304800)
7: (-0.70041203020487 1.34701081255472 0.20474228427596)
8: ( 0 0 5.00000000000000)

The last one was the one I used for constructing the example.

Answer 2

If you know the equations of the three arcs, let P(x,y) be a point inside the region bounded by the three arcs. Then find the three expressions for the distance from P to each of the arcs. The point which makes these three expressions equal will be the center of your circle, and the radius will be the distance to an arc.

Answer 3

Is it possible to draw tangent lines to each of the arcs (convex)
or connecting endpoints (concave)
and then inscribe a circle inside the resulting triangle?

Steve