Area of Intersection of Two Circles

Date: 12/1/95 at 20:44:45
From: Anonymous
Subject: Area of intersection of circles

My teenage son asked me for the formula for the area of 
intersection of two arbitrary circles,

center (x0,y0), radius r0
center (x1,y1), radius r1

I do not know the answer.  Can you help me?

Thanks,
Jack

Date: 5/31/96 at 11:1:26
From: Doctor Anthony
Subject: Re: Area of intersection of circles

I will describe a figure which you should draw out, and then follow 
my working making reference to the figure.  

Let A be the center of the circle (x0,y0) and B be the center of the 
other circle (x1,y1).

Draw the circles with appropriate radii r0 and r1 so that there is a 
reasonable amount of overlap.  The length AB is calculated from the
coordinates of the centers:

  AB = sqrt{(x1-x0)^2 + (y1-y0)^2}

For convenience let this length be denoted by c.

The two circles intersect in two points which I will label C and D.
Now we must calculate the angles CAD and CBD, and we do this 
using the cosine formula. In fact it is half of these angles that we first 
calculate, using triangle CAB.

      r0^2 = r1^2 + c^2 - 2*r1*c*cos(CBA)
           .
           .
  cos(CBA) = (r1^2 + c^2 - r0^2)/(2*r1*c)

Having found CBA, then CBD = 2(CBA).

Similarly,

  cos(CAB) = (r0^2 + c^2 - r1^2)/(2*r0*c)

and then   CAD = 2(CAB)
 
Express CBD and CAD in radian measure. Then we find the segment
of each of the circles cut off by the chord CD, by taking the area of 
the sector of the circle BCD and subtracting the area of triangle BCD.  
Similarly we find the area of the sector ACD and subtract the area of 
triangle ACD.

  Area = (1/2)(CBD)r1^2 - (1/2)r1^2*sin(CBD)
       + (1/2)(CAD)r0^2 - (1/2)r0^2*sin(CAD)

Remember that for the area of the sectors you must have CBD and 
CAD in radians.

-Doctor Anthony
 The Math Forum

Date: 5/31/96 at 12:1:26
From: Doctor Alex
Subject: Re: area of intersection of circles

I would like to offer a tip. 

Dr Anthony's final formula is a general solution. However, if the two 
circles are of the SAME radius please note that the area is symmetrical 
about the chord CD.

Therefore, you only need to find the area in one half of the 
intersection and multiply by 2.

A shorter equation is Area =2*( (1/2)(CBD)r1^2 - (1/2)r1^2.sin(CBD)).

-Doctor Alex
 The Math Forum

Date: 06/06/2003 at 21:05:53
From: Pete Krell
Subject: Your site helped me! But I'd like to enhance your answer...

I am a business professional and ex-math major.  I was trying to 
calculate the percentage of overlap of two coverage areas, e.g. the 
areas of intersection of two circles. Your answer and derivation 
gave me what I needed - thank you. 

But many people trying to solve this problem do not need to deal 
with x and y coordinates. And if the radii are equal, the formula 
is much simpler even than the semi-simplification the second professor 
added.  

I believe that:

cos([CBA]) = c^2 / 2rc = c/2r
Let q = [CBD] = 2*[CBA] = 2*acos(c/2r)
As Professor 2 said, Area = 2(qr^2/2 - sin(q)r^2/2),
but this simplifies easily.

So the whole thing boils down to:
Area = r^2*(q - sin(q))  where q = 2*acos(c/2r),
where c = distance between centers and r is the common radius.

Again, this was the best source I could find. (I found one 
other that was quite wrong.) Thought it might help someone else.