Area of Intersection of Two Circles

center (x0,y0), radius r0

center (x1,y1), radius r1

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.

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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)).

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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 s

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.