Can someone make this a little clearer Actually, you can get the general case wi
ID: 3094711 • Letter: C
Question
Can someone make this a little clearer Actually, you can get the general case without calculus,too.Still assume both circles have radius 1, but allow the distancebetween their centers to be variable--d.
For this one I will name the points. A is one point of intersectionand B is the other. O is one circle center and P is the other. Drawsegments AB, OP, AO, AP, BO, and BP. The intersection between ABand OP is point C.
The first step is to find the area of triangle OAB. Its height isd/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 isto find the area of the wedge enclosed by the angle AOB. The angleis 2 * arccos (d/2), so (taking the angle in radians) the arc areais 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; doublethat to find the whole area of intersection. Can someone make this a little clearer Actually, you can get the general case without calculus,too.
Still assume both circles have radius 1, but allow the distancebetween their centers to be variable--d.
For this one I will name the points. A is one point of intersectionand B is the other. O is one circle center and P is the other. Drawsegments AB, OP, AO, AP, BO, and BP. The intersection between ABand OP is point C.
The first step is to find the area of triangle OAB. Its height isd/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 isto find the area of the wedge enclosed by the angle AOB. The angleis 2 * arccos (d/2), so (taking the angle in radians) the arc areais 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; doublethat to find the whole area of intersection.
Explanation / Answer
For at least the special case where the center of one circle fallson the edge of the other circle, you don't have to use calculus; alittle figuring gets the answer. Draw a vertical line between the top intersection point and the onebelow it, thus dividing the region of the two circles combined inhalf. I say this here so as not to break up the explanationlater. The quadrilateral formed by the 2 centers of the circles and theintersection points of the edges is composed of 2 equilateraltriangles stacked on one another (all edges = radius of circle). Sothe 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 inthe same manner. If you draw a line the way I described in the lastparagraph for each of these other circles too, the three lines forman equilaterial triangle within the very first circle. If you drawthe figure as I described and know the side proportions of a30-60-90 triangle, you can see that the length of a side of thistriangle is 3 ^ .5 radii. The height from one of these sides to theopposite point is clearly 1.5 radii. So the area of the wholetriangle is 3 ^ 1.5 / 4. The area of the whole circle is pi, so thearea 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. You can find the equatation of the part of the circle that enclosesthe intersection. Something like y=sqrt(r²-x²)-a. Integrate it, multiply with 2 (if the 2 circles have the sameradius), the result is the area.