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Calculating area of intersecting circles

  1. Aug 23, 2004 #1
    Hi All,

    I went to this website where someone wants to know how to calculate the area of some intersecting circles in this diagram on the page. I took Geometry and Trig but unfortunately my brain turned to mush over the summer and I can't remember how to do this, can someone here tell me how I might go about calculating this?

    Here's the link to the page: http://butchlafonte.50megs.com/photo3.html

    Thanks,
    Jason O
     
  2. jcsd
  3. Aug 23, 2004 #2
    You can find the equatation of the part of the circle that encloses the intersection. Something like y=sqrt(r²-x²)-a.
    Integrate it, multiply with 2 (if the 2 circles have the same radius), the result is the area.

    I added an image to explain it better.
     

    Attached Files:

  4. Aug 23, 2004 #3
    For at least the special case where the center of one circle falls on the edge of the other circle, you don't have to use calculus; a little figuring gets the answer.

    Draw a vertical line between the top intersection point and the one below it, thus dividing the region of the two circles combined in half. I say this here so as not to break up the explanation later.

    The quadrilateral formed by the 2 centers of the circles and the intersection points of the edges is composed of 2 equilateral triangles stacked on one another (all edges = radius of circle). So the angle between one intersection point and the other is 60 + 60 = 120, so 3 circles can be fit around the edge of the first one in the same manner. If you draw a line the way I described in the last paragraph for each of these other circles too, the three lines form an equilaterial triangle within the very first circle. If you draw the figure as I described and know the side proportions of a 30-60-90 triangle, you can see that the length of a side of this triangle is 3 ^ .5 radii. The height from one of these sides to the opposite point is clearly 1.5 radii. So the area of the whole triangle is 3 ^ 1.5 / 4. The area of the whole circle is pi, so the area of all the 3 edge slices outside the triangle together is pi - 3 ^ 1.5 / 4. So the area of each of these edge slices is pi / 3 - 3 ^ .5 / 4. Double that and you have the region you are looking for, the shared region between 2 circles: its area is 2 * pi / 3 - 3 ^ .5 / 2.
     
  5. Aug 24, 2004 #4
    Actually, you can get the general case without calculus, too.

    Still assume both circles have radius 1, but allow the distance between their centers to be variable--d.

    For this one I will name the points. A is one point of intersection and B is the other. O is one circle center and P is the other. Draw segments AB, OP, AO, AP, BO, and BP. The intersection between AB and OP is point C.

    The first step is to find the area of triangle OAB. Its height is d/2. By the pythagorean formula, its base is 2 * (1 - d ^ 2 / 4) ^ .5, so its area is d / 2 * (1 - d ^ 2 / 4) ^ .5. The next step is to find the area of the wedge enclosed by the angle AOB. The angle is 2 * arccos (d/2), so (taking the angle in radians) the arc area is pi * 2 * arccos (d/2) / (2 * pi) or just arccos (d/2). So arccos (d/2) - d / 2 * (1 - d ^ 2 / 4) ^ .5 is one of the slivers; double that to find the whole area of intersection.
     
  6. Aug 27, 2004 #5
    Thank you both for the help. I very much appreaciate it :smile: . I still have to sit down and follow the instructions you posted but as I read it I did understand what you were explaining. Again thank you both.

    - Jason O
     
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