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Area of Intersection of Two Circular Segments

Date: 04/20/2007 at 03:45:14
From: Claudio
Subject: area of two intersecting circular segments

Given a circle of radius r and center c, suppose two intersecting
chords AB and CD (intersecting in P) form two circular segments.  How
do I compute the area of the intersection of the two circular segments?

I can easily find out the area of the whole circle and of the two
circular segments, but I'm having trouble with the area of the
intersection.  Here's what I think might work:

W = center of the circle
r = radius of the circle
AB = first chord
CD = second chord
P = point of intersection of the two cords
a = height of the circular segment created by AB
b = height of the circular segment created by CD
K = area of the overlapping section

PB = sqrt( r^2 - (r-a)^2 ) 
CP = sqrt( r^2 - (r-b)^2 )
chord CB = sqrt( PB^2 + CP^2 )

X = area of triangle BCP = PB*CP/2

h = height of the triangle WCB = sqer( r^2 - (CB/2)^2)
Y = area of the triangle WCB = h*CB/2

theta = angle of the circular section created by the arc from C to B =

Z = the area of the circular section created by the arc from C to B = 

K = Z - Y + X



Date: 04/23/2007 at 09:28:36
From: Doctor George
Subject: Re: area of two intersecting circular segments

Hi Claudio,

Thanks for writing to Doctor Math.

It looks like your analysis is using the Pythagorean theorem with
triangles that are not necessarily right triangles.  Let me show you
what I came up with, then you can compare your results with mine.

Problem Statement
Consider a circle with center O and radius R.  Construct chords AB and
CD that intersect at point P such that O is not in the region bounded
by segments AP, PD and arc AD.  Find the area of that region, assuming
that the lengths of CA, AD, and DB are known.

To simplify notation define the following.

  a = angle COA (in radians)
  b = angle AOD
  c = angle DOB

The solution can be written as a function of R, a, b and c.

We can relate the angles to the known segment lengths like this.

  CA = 2R sin(a/2)
  AD = 2R sin(b/2)                                 (1)
  DB = 2R sin(c/2)

A known theorem about circles tells us that

  angle CDA = a/2
  angle DAB = c/2

Looking at triangle APD we can say

  angle APD = pi - (a+c)/2

From the law of sines we can write

  sin[pi - (a+c)/2]   sin(c/2)
  ----------------- = --------                    (2)
          AD              PD

The area of triangle APD is

  AD/2 * [PD * sin(a/2)]

If we use equations (1) and (2) to substitute for AD and PD we can
write the area of triangle APD as

       sin(a/2) sin^2(b/2) sin(c/2)
  R^2 ------------------------------

Now we just need to add the area bounded by chord AD and arc AD. This
area is

  R^2 (b - sin(b))

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum 

Date: 04/25/2007 at 02:54:04
From: Claudio
Subject: Thank you (area of two intersecting circular segments)

Dear Dr. George,

Thanks very much for your help.  The solution you've found is very
helpful to me and I really appreciate your time.



Date: 02/21/2016 at 05:57:14
From: Graeme 
Subject: the area of intersection of two circular segments 

The total area of intersection was provided above. But I am rather looking 
for the area of the intersection of the two segments.

I looked at that other solution, but did not find it strictly relevant.

It has been 40+ years since I did this sort of stuff. I'm having 
difficulty just setting up the relevant methodology ...

Date: 02/22/2016 at 11:38:41
From: Doctor Rick
Subject: Re: the area of intersection of two circular segments 

Hi, Graeme.

As far as I can see, what Doctor George found there *was* the area of 
overlap between two segments of the circle. He did not, however, state 
clearly the order of the points around the circle: they are not in the 
usual alphabetized A, B, C, D!

Here is a figure that I believe represents what Doctor George 
was describing:


- Doctor Rick, The Math Forum
Associated Topics:
High School Euclidean/Plane Geometry
High School Trigonometry

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