Intersecting Circles Problem

[Fringe Pioneer]

Imagine two congruent circles of radius α (from Greek "aktina" = "radius"). The centers of the circles are colinear with the origin (0, 0) of a Cartesian plane in a way such that the origin is the midpoint of the segment formed by the ...circles' centers. The distance κ (from Greek "katina" = "center") from the center of a circle to the origin is less than the radius α. When graphed, the arbitrary result should resemble a Venn Diagram.

Given the above, at what arbitrary value of κ will the area of the circles' intersection equal the sum of the non-intersecting parts of the circles?

To simplify the math, I put the circles on the x-axis. Doing that, the center of one circle should be (-κ, 0) and the center of the other circle should be (κ, 0).

I then formed four triangles by drawing a segment from (-κ, 0) to (κ, 0), drawing two segments from (-κ, 0) to both of the points of intersection, and drawing two segments from (κ, 0) to both of the points of intersection. I looked at one of the four small triangles, noting its horizontal leg to be of length κ and its hypotenuse to be of length α. Since I would be taking an integral with polar coordinates, I needed to find the angle formed by the hypotenuse and the horizontal leg. This angle I found to be arccos(κ/α). Since I needed to find the angle from one point of intersection to the other point of intersection, and since all four small triangles were congruent, I doubled the angle and got 2*arccos(κ/α).

I proceded by integrating the slice of the circle from one point of intersection to another point of intersection. Since polar integrals of circles are much easier than Cartesian integrals of circles, I took a double integral from angle (π - arccos(κ/α)) to angle (π + arccos(κ/α)) and from radius 0 to α of the function R. integral(integral(R, 0, α), π - arccos(κ/α), π + arccos(κ/α)) This evaluated to integral(α^2/2, π - arccos(κ/α), π + arccos(κ/α)). This evaluated to (α^2)*arccos(κ/α), giving me the area of the "slice" of a single circle.

I went back to the isoscelese triangle and used the Pythagorean Theorem to find its area, which was κ*(α^2 - κ^2)^(1/2).

To find the area of half the intersection of the circles, I subtracted the area of the triangle from the area of the entire "slice." I then doubled it to find the area of the intersection of the circles. This came out as 2(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2)).

The area of each non-intersecting part of a circle was simply the whole circle minus the area of the intersection. This came out as π(α^2) - 2(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2)). Doubling this to get the combined area of both non-intersecting parts of the circles, I got 2π(α^2) - 4(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2)).

I set the area of the non-intersecting parts equal to the area of the intersection.

2π(α^2) - 4(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2)) = 2(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2))

I added the second term of the first part of the equation to both sides of the equation, as follows.

2π(α^2) = 6(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2))

I factored out 2 from both sides.

π(α^2) = 3(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2))

I divided both sides by 3 and by α^2.

π/3 = 2arccos(κ/α) - κ*(α^-2)*(α^2 - κ^2)^(1/2)

After that, I got stuck trying to manipulate the equation for κ on account of the arc cosine and the square root. Please help me solve for κ.

[tzetze]

To find the area of half the intersection of the circles, I subtracted the area of the triangle from the area of the entire "slice." I then doubled it to find the area of the intersection of the circles. This came out as 2(2(α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2)).

There is a "2" too much. You should have:
2((α^2)*arccos(κ/α) - κ*(α^2 - κ^2)^(1/2))

π/3 = 2arccos(κ/α) - κ*(α^-2)*(α^2 - κ^2)^(1/2)

The preceding error means that you should have:
π/3 = arccos(κ/α) - κ*(α^-2)*(α^2 - κ^2)^(1/2)

Put arccos(k/a) = T , the angle that you have.
cos(T)= k/a
sin(T)= sqrt(1-(k²/a²))

You can then rewrite your equation as:

Pi/3 = T - cos(T)*sin(T)
2*Pi/3 = 2T - 2*cos(T)*sin(T)
2*Pi/3 = 2T - sin(2T)
sin(2T) = 2T - 2*Pi/3

This equation, in the unknown quantity x=2T, cannot be solved
algebrically, like the equation cos(x)=x.

But you can plot the graph of
y=sin(x)
and the graph of
y=x-2*Pi/3
and see that there is only one intersection point, which gives x.
Then you get T and k/a.

Hope this helps.

[Fringe Pioneer]

Yes, that helps immensely. I took out the extraneous 2s (which I may have placed in because I may have forgotten about the common factor of 1/2 while integrating) and observed what I had. Then I looked at your note to take apart (κ*α^-2)(α^2 - κ^2)^(1/2) as κ/α * (α^2 - κ^2)^(1/2)/α. After reading on what you did concerning finding the sine of the arccosine of κ/α and the double angle property, I saw the genius of your manipulations.

Essentially, I just went through your steps and found the solution to the system of equations for sin(x) = x - 2π/3, where x = 2arccos(κ/α).

I eventually found that, in order for the intersection of the circles to equal the non-intersecting areas of the circles, κ must be approximate to 0.265α.