Overlapping area of 2 circles (II)

[Guest]

Inside a circle E with radius 2 units , there are 2 circles A and B
both with radius 1 unit . A touches E at a point and keep fixed ,
while B rolls randomly along the circumference of E .
Find the expectation of the overlapping area of A and B .

[shyamjayakannan]

Screenshot 2026-03-17 195319.png (86.74 KiB) Viewed 78 times

Consider the situation as shown in the image above. The large circle has a radius of 2 and both smaller circles have radii of 1. Let the second circle be at an angle [tex]\theta[/tex] from the first one. In this situation, the overlapping area can be found as 2 x area of the segment in the overlapping area. Now, to find the area of the segment, we use: [tex]ar(\text{segment})=ar(\text{sector} AFG)-ar(\triangle AFG)[/tex]

Now, [tex]\angle AGF=180^\circ-2\angle AFG=\pi-2\times\frac{\theta}{2}[\text{in rad}]=\pi-\theta[/tex]. So, we have:

[tex]ar(\text{segment})=\pi r^2\frac{\angle AGF}{2\pi}-\frac{1}{2}AG\times FG\sin{\angle AGF}=\frac{\pi-\theta-\sin{(180^\circ-\theta)}}{2}=\frac{\pi-\theta-\sin\theta}{2}[/tex] and hence the total overlapping area is twice of this = [tex]\pi-\theta-\sin\theta[/tex]. Note that this is only for [tex]0\le\theta\le\pi[/tex] and for [tex]\pi<\theta\le2\pi[/tex] the expression would become [tex]\pi-(2\pi-\theta)-\sin(2\pi-\theta)=\theta-\pi+\sin\theta[/tex]. Combining both expressions into 1 for [tex]0\le\theta\le2\pi[/tex], this becomes = [tex]|\pi-\theta|-|\sin\theta|[/tex]

Now, let us divide the [tex]2\pi[/tex]rad into [tex]n[/tex] equal parts and assume that the second circle can only rotate in discrete angles that are multiples of [tex]\frac{2\pi}{n}[/tex]rad. This will make computing the probability easier. So, the formula for the expected overlapping area becomes:

[tex]E_n=\sum(\text{probability of the second circle to be at angle }\theta)\times(\text{overlapping area at angle }\theta)=\sum_{k=1}^n\frac{1}{n}\times\left(\left|\pi-\frac{2k\pi}{n}\right|-\left|\sin\frac{2k\pi}{n}\right|\right)[/tex]

Now, for the actual situation we need to find [tex]\lim_{n\to\infty}E_n=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\left(\left|\pi-\frac{2k\pi}{n}\right|-\left|\sin\frac{2k\pi}{n}\right|\right)=\int_0^1(|\pi-2\pi x|-|\sin{2\pi x}|)\ dx[/tex]

[tex]=\int_0^\frac{1}{2}(\pi-2\pi x-\sin{2\pi x})\ dx+\int_\frac{1}{2}^1(2\pi x-\pi+\sin{2\pi x})\ dx=\boxed{\frac{\pi^2-4}{2\pi}}[/tex]