Find the area between two circles if the circumference of one circle passes through the origin of the other circle and vice versa. (See the figure below.)
In the diagram to the left, the graphs of and are shown and the overlap is shaded. Find area of the shaded region.
To make things easier, let’s just find the area of half of the intersection. Let’s do that by first drawing a line down the middle of the intersection.
Now, let’s look at the left circle and inscribe an equilateral triangle.
From the construction of the triangle in the circle, it is evident that the center of the triangle is at the origin. Now, let’s bisect the angles of the equilateral and notice that the lines all meet in the middle of the circle, ie the origin. As is shown in the next picture.
These new lines are labeled red in this diagram. The lines themselves have radius 1 since they come from the origin and reach the circumference of the circle given by the equation . What’s more, the lines partition the circle into three equal sectors.
Let us just consider the sector of the circle that is contained between the two red lines on the right side of the circle:
Now, we can find the area of the triangle bounded by the two red lines and the vertical line. Since we bisected the angles of the equilateral triangle, we know the top and bottom angles must be half of 60º, namely 30º. That makes the angle in between the two red lines 120º. Also, we know that the length of each of the red lines is 1 since they are both radiuses from the circle . All of this is labeled in the next picture:
Using the Law of Sines, we can find the length of the side opposite the 120º angle. We have the following equation:
We can now find the area of the triangle. The triangle has “height” 0.5 (on the x-axis) and base (on the y-axis). So, the area will be
Alternatively, we could have used the formula , where a and b are both 1 and the angle is 120º.
Regardless, we are almost done. Let us find the area in the sector of the circle that is not contained in the triangle. This area will be half of the area of the intersection of the two curves. A picture will help again. We are looking for the area of the shaded region.
We know that the area of the sector will be one third of the circle (since the sector has an angle 120º and there are 360º in a circle.) Since the circle has radius 1, the area of the sector will be . We found the area of the triangle to be . Therefore, the area of the shaded region is .
Recall, we are looking for two of these shaded regions. So, the area of the intersection of the two circles will be .
In the above example, we used radius 1. If we generalize to any radius, r, for both circles (and the centers are moved accordingly), then the area can be found by the following formula: