Area of Intersection of Two CirclesDate: 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. |
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