Does anyone know how to find the area between three tangent circles? I have drawn a picture. Assume that the circles have radii of 2, 3 and 4 (obviously they aren’t to scale in my diagram) and they are tangent wherever they touch. How do I find the area of the yellow space?
Thanks!
–FCOD
One obvious way would be to find the area of the triangle that’s defined by their center points, plus the area in each circle outside that triangle, minus the area of the three circles.
ETA: I’m not sayin’ it’s necessarily easy to do this.
I thought of that but I don’t know how to find the area of the circle minus the triangle either lol.
–FCOD
I haven’t tried to do this, and it might be harder than it sounds, but…
If you know where the circles contact each other, then (using circle 3 as an example) you can get the angle from contact of 2 to the contact of 4, and therefore the arc length, s, of that segment (s=r*theta).
You could then make an approximation using those arc lengths as if they were lengths of a triangle, but I guess that’s not what you want.
You could find a function for each of those three arc-lengths, and determine the area by taking the integral between two curves (you’d have a discontinuity at the connecting point of the two “top” curves)
The area of the circle that coincides with the triangle would be easy to figure out using the angle of the vertex within that circle. The area of the pie piece would be equal to that angle divided by 360 times the area of the circle. To figure out the angle, just use basic trig; you already know the length of each edge of the triangle.
The general algorithm:
E: On first thought I was assuming the sides of the triangle would be the sum of the circle radii, but picturing circles with widely varying radii, I see this is incorrect. I think that throws a wrench in my idea.
You know the lengths of the three sides of the triangle (formed by the circles’ centers) - that should be enough to get you all three angles. Once you’ve got that:
Finding the area of the triangle should be straightforward. Subtract from that the area included in each circle - which should be easy because you know each angle and radius - and you’re done.
This may be harder than it sounds - obscure trig rules that I don’t know by heart may come in very handy. I’m going to start working on this…
ETA: In other words, what Apex Rogers said…
Finding the area of a triangle with corners at the three centers should be pretty easy. Find the angles that exist at each corner. The area of that circle that is cut by the angle is proportional to the area of the whole circle. (Area of wedge)/(Area of circle)=(angle)/(360).
Draw the triangle that connects the three circle centers.
Drop a perpendicular from one enter to the far side of the triangle.
Set up a quadratic sum to determine the length of all three sides of one of these sub-triangles.
subtract to get the remaining length of the other triangle side
Use the lengths of the sides to set up a sine, cosine or tangent relationship to determine the two partial angles at the center of the circle you dropped a construction line from. This will necessariy include some inverse trigonometric function.
Once you have the angles, you can find the area of the “wedge” as a fraction of the circle’s area. (e.g. 2pirr90/360, or simply rr*theta if you’re using radians )
Use the general form of your solution for the other two wedges, and subtract the three from the area of the triangle.
A similar problem is solved here, and may help visualize the process.
How is that incorrect? Each side must pass through a point of tangency, and therefore consists of the radius of one circle end-to-end with the radius of the other circle.
And since everyone’s danced around the question, the trig relationship you’d use to find the angles is the Law of Cosines.
This triangle calculator gives the angles of the the 5,6,7 triangle as 44.4, 57.1 and 78.5.
Heron’s formula [triangle area = sqrt (s * (s-a) * (s-b) * s-c)) where s = (a + b + c) / 2 ] gives the triangle’s area as 14.697.
Further simple calculations yield a result of 1.28 square units.
(Which seems a bit small, but that’s what my feeble late-evening abilities twice produced).
Yeah I had a moment of panic after submitting and “corrected” myself after a bit of second-guessing. Having thought it out fully I realize my first instinct was correct.
The others have given the process for doing so. Me, I thought I’d check Xema’s work, so, well, I cheated some. I drew up the problem in AutoCAD, a computer drafting program, drew what’s called a polyline around the area in question, and the computer says the area is 1.2702 square units. Close enough.
But yes, my method would otherwise be to figure the angles for the triangle connecting the circle centers, find the area of that triangle, then use the angles to subtract out the segment areas.
Alright, let’s see:
Get the area of the triangle that forms between their center points. That part should be easy enough, yes? Using your example of the 2, 3, and 4 radii, we can get a triangle with leg lengths of 5, 6, and 7.
We use Heron’s formula (i admit I had to look up the name, but I did know it existed!) which says: Area = sqrt(s(s-a)(s-b)(s-c))
In this case, a, b, and c are the leg lengths, and s is 1/2(a+b+c). So, doing all that math, the area of the triangle is 14.69 (heh…69)
Next we need to find the area of the arc segments of each circle contained within the triangle. For this, we need each angle in the triangle. I’m sure there’s a simpler way to do this, but the best way I know is the law of sines:
a/sin A = b/sin B = c/sin C = 2R.
R is the radius of the circumcircle, so 2R is the diameter. As you might guess, the circumcircle is the circle formed around the triangle with the three points on the circle. The diameter of the circumcircle is:
abc/2*Area. We know all of these, so the diameter is 7.15.
Finally, we can get the angles. The angle named for each leg is the one opposite that leg, so first up:
a/sin A = 7.15 = 5/sin A.
Arcsin A = 5/7.15 = 44.37 degrees. Since length a was the smallest length, it was opposite the angle formed in the “4” triangle.
b/sin B = 7.15 = 6/Sin B
Arcsin B = 6/7.15 = 57.05 degrees, which is the angle in the “3” triangle.
c/sin C = 7.15 = 7/sin C
Arcsin C = 7/7.15 = 78.24 degrees.
Let me check that those angles add up to 180…yup (well, close enough since I rounded.)
So, now we can get the area of the arc segments.
Circle “2”: Area = 4Pi, arc segment area = 4Pi(78.24/360) = 0.87Pi
Circle “3”: Area = 9Pi, arc segment area = 9Pi(57.05/360) = 1.43Pi
Circle “4”: Area = 16Pi, arc segment area = 16Pi(44.37/360) = 1.97Pi
So, the total area of all the arc segments is 4.27Pi = 13.41.
From before, the area of the triangle is 14.69.
14.69-13.41 = 1.28
1.28
I pray I did all my math right…
Oh, and only now do I see there’s an internet program called the triangle calculator that could have done all this stupid math FOR me.
Damn it! i did all that work!
Edit: oh, and the law of cosines would probably have been easier than the law of sines and the circumcircle.
If you want to use an alternate method just for shit and giggles, you can also try using the circle segment area calculation to subtract out just the portions of circle that lie within the triangle defined by the points of contact.
(You’ll still need to use the law of cosines, though.)
I’m not sure I used the right terminology, then. As others have suggested and done, subtract out the portion of each circle (whatever such a portion is called) within the triangle connecting the circle centers.
There’s a pretty straightforward Monte Carlo algorithm for this problem that generalizes to any number of pairwise tangent circles. The key observation is that the line segment between the centers of any two tangent circles passes through the point of tangency. Therefore, the shape whose area we’re trying to estimate lies in a convex polygon defined by the centers. And in a convex polygon, every point can be represented as a convex combination of the vertices.
Given n > 2 pairwise tangent circles, the algorithm is as follows: generate n random numbers from a uniform distribution on [0, 1] and divide each of them by their sum. Form a convex combination of the centers using those numbers as your weights. Check to see if your new point falls inside any of the circles; if not, it must fall inside the shape we’re interested in. After a sufficient number of iterations, you have a pretty good estimate of the ratio of the area of the region of interest to the area of the polygon, so you can just multiply by the area of the polygon to get your final answer.
The only slightly tricky part in here is that you have to know how to calculate the area of a polygon given its vertices. I suggest looking here.
On reflection, the Monte Carlo algorithm is overkill here. You know the area of the polygon defined by the centers of the circles, and you know the area of each circle’s intersection with the polygon[sup]1[/sup], so it’s a simple matter of subtracting one from the other. Still, it’s nice to know that any time you have a shape bounded by a polygon, you can estimate the area to arbitrary accuracy using a little randomness.
[sup]1[/sup]: For each center, subtract the adjacent centers to get a vector. Take their dot product, divide by their norms, and that gives you the cosine of the angle defined by those two vectors. Multiply that angle by the radius of the circle, and there’s your area.
A closed-form solution is decidely useless. 
A = (1/2) { (r[sub]A[/sub] + r[sub]B[/sub] ) (r[sub]A[/sub] + r[sub]C[/sub] ) ( 1 - { [r[sub]A[/sub][sup]2[/sup]+2r[sub]A/sub+r[sub]B[/sub]r[sub]C[/sub]]/[(r[sub]A[/sub]+r[sub]B[/sub])(r[sub]A[/sub]+r[sub]C[/sub])] }[sup]2[/sup] )sup[/sup] - [ r[sub]A[/sub][sup]2[/sup]cos[sup]-1/sup + r[sub]B[/sub][sup]2[/sup]cos[sup]-1/sup + r[sub]C[/sub][sup]2[/sup]cos[sup]-1/sup ] }
If you squint, you can see a term for the area of the big triangle and three negative terms for the areas of the sectors in the circles. The first is just (1/2)baseheight (with law of cosines (or, equivalently, the vector dot product) making an appearance) and the latter three terms are just -([symbol]p[/symbol]r[sup]2[/sup])*(angle/2[symbol]p[/symbol]), since the area of the sector is the area of the circle times the fraction of the circle inside the sector.
I tried keying that into my calculator to work out the OP’s example, but my fingers fell off before I got halfway through it. Thanks a lot, Pasta. :mad:
Now, where’s that left pinkie?
