Finding the Area of Overlapping CirclesDate: 11/08/2004 at 21:31:10 From: Nick Subject: Area of Overlaping Circles Given two circles: circle O: (x - 1)^2 + (y - 2)^2 = 25 circle P: (x - 4)^2 + y^2 = 16 What is the area of the overlapping part of the circles? I couldn't find anything on how to do this kind of problem, and I'm stuck. Date: 11/12/2004 at 11:18:59 From: Doctor Cristian Subject: Re: Area of Overlaping Circles Hi Nick, and thanks for asking Dr. Math! I can think of no other way than the straight-forward one, but when I try to put it all together for the general case, I found that there's no "nice" answer, so I can't give you a formula to help with other similar cases. So bear with me, and we'll be getting to the answer you're looking for. First, note that the relative position of the two circles (and thus the area of their intersection) doesn't change upon moving and rotating the coordinate system. So we can say that one circle is centered in the origin, and the other one is centered at (d, 0), where d is the original distance between the two centers. This is purely for easing our work, since we could, in theory, work with the given data. The actual value for d, in your case, is d = sqrt( (x1-x2)^2 + (y1-y2)^2 ) = sqrt( (1-4)^2 + (2-0)^2 ) = sqrt( 9 + 4 ) = sqrt(13) So now that our picture looks a little better (you can look at http://mathforum.org/dr.math/gifs/nick.11.12.04.jpg to see a sketch of it), we should find the coordinates of the two intersection points. That can be done by solving the system of the equations of the circles: x^2 + y^2 = r1^2 (x-d)^2 + y^2 = r2^2 Substracting the two, we get -2dx + d^2 = r2^2 - r1^2, or x = (r1^2 + d^2 - r2^2) / (2d), so y^2 = r1^2 - x^2 = 2 r1^2 d^2 + 2 r2^2 d^2 + 2 r1^2 r2^2 - r1^4 - r2^4 - d^4 --------------------------------------------------------- 4 d^2 which gives us the two solutions (one positive and one negative): yp = sqrt( ... ), and yn = -sqrt( ... ) In your case, these are x = (25 + 13 - 16) / (2sqrt(13)) = 11 sqrt(13) / 13 yp = sqrt( 25 - (121/13) ) = sqrt(204/13) = 2 sqrt(51/13) = 2 sqrt(663) / 13 yn = -yp = -2 sqrt(663) / 13 Now we can find the angles that these chords (it is only one segment, but it is a *different* chord in each circle) subtend. angle = 2 arcsin( (length/2) / radius ), where length is length = yp - yn In the first circle, a1 = 2 arcsin ( (2 sqrt(663) / 13) / 5 ) = 2 arcsin ( 2 sqrt(663) / 65 ), which approximates to ~ 2 * 0.9145 = 1.829 (radians), which is about 105 degrees--this can be checked in the picture. In the second circle, a2 = 2 arcsin ( (2 sqrt(663) / 13) / 4 ) = 2 arcsin ( sqrt(663) / 26 ), which approximates to ~ 2 * 1.4317 = 2.8634 (radians), which is about 164 degrees--this too can be checked in the pic. And now the two segments of the circles come into play. You can review what they are and a few formulas about them at Circle Formulas http://mathforum.org/dr.math/faq/formulas/faq.circle.html#segment The area is simply the area of the sector of circle minus the area of the triangle. Keep in mind though that the segment of circle that lies closer to the center of one of the two circles actually belongs to the *other* circle! The area of the segment belonging to the first circle is A_seg_1 = r1^2 a1 / 2 = 25 arcsin ( 2 sqrt(663) / 65 ) The area of the corresponding triangle is A_tri_1 = x * yp = (11 / sqrt(13)) * (2 sqrt(51/13)) = 22 sqrt(51) / 13 So the area of the righternmost half of the intersection is A_sect_1 = A_seg_1 - A_tri_1 = 25 arcsin ( 2 sqrt(663) / 65 ) - 22 sqrt(51) / 13 Similarly, the other three areas are A_seg_2 = r2^2 a2 / 2 = 16 arcsin ( sqrt(663) / 26 ) A_tri_2 = (d - x) * yp = (2 / sqrt(13)) * (2 sqrt(51/13)) = 4 sqrt(51) / 13 A_sect_2 = A_seg_2 - A_tri_2 = 16 arcsin ( sqrt(663) / 26 ) - 4 sqrt(51) / 13 And, finally, we've found the answer of A_intersection = A_sect_1 + A_sect_2 = 25 arcsin ( 2 sqrt(663) / 65 ) - 22 sqrt(51) / 13 + 16 arcsin ( sqrt(663) / 26 ) - 4 sqrt(51) / 13 ~ 25 * 0.9145 + 16 * 1.4317 - 2 * 7.1414 = 31.487 Write back if you have any questions about this long and awkward-looking process. - Doctor Cristian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/