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 161 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]

[wiskunde]

Let the large circle \(E\) have center \(O\).

Both \(A\) and \(B\) have radius \(1\), and because each is inside the circle of radius \(2\), their centers lie on the circle of radius \(1\) around \(O\).

Fix \(A\). Let the angle between the centers of \(A\) and \(B\) be \(\theta\). Since \(B\) rolls uniformly around the circumference,

$$ \theta\sim \text{Uniform}(0,2\pi). $$

The distance between the centers is therefore

$$ d=2\sin\frac{\theta}{2}. $$

For two unit circles whose centers are distance \(d\) apart, their overlapping area is

$$ L(d)=2\cos^{-1}\left(\frac d2\right) -\frac d2\sqrt{4-d^2}. $$

Substitute \(d=2\sin(\theta/2)\). Put \(x=\theta/2\):

$$ L=2\cos^{-1}(\sin x)-2\sin x|\cos x|. $$

Hence the expected overlap is

$$ E[L] =\frac1\pi\int_0^\pi \left[2\cos^{-1}(\sin x)-2\sin x|\cos x|\right]dx. $$

Now,

$$ \int_0^\pi \cos^{-1}(\sin x)\,dx =\frac{\pi^2}{4}, $$

and

$$ \int_0^\pi \sin x|\cos x|\,dx=1. $$

Therefore,

$$ E[L] =\frac1\pi\left(\frac{\pi^2}{2}-2\right). $$

So the answer is

$$ \boxed{\frac{\pi}{2}-\frac{2}{\pi}} $$

square units.

Numerically,

$$ \boxed{E[L]\approx0.93418\text{ square units}}. $$

So the expected overlapping area is approximately \(0.934\) square units. For more information you can also check https://wiskundehulpnl.com/

[nycmath]

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 .

This is advanced geometry above my current skills.

[nycmath]

Let the large circle \(E\) have center \(O\).

Both \(A\) and \(B\) have radius \(1\), and because each is inside the circle of radius \(2\), their centers lie on the circle of radius \(1\) around \(O\).

Fix \(A\). Let the angle between the centers of \(A\) and \(B\) be \(\theta\). Since \(B\) rolls uniformly around the circumference,

$$ \theta\sim \text{Uniform}(0,2\pi). $$

The distance between the centers is therefore

$$ d=2\sin\frac{\theta}{2}. $$

For two unit circles whose centers are distance \(d\) apart, their overlapping area is

$$ L(d)=2\cos^{-1}\left(\frac d2\right) -\frac d2\sqrt{4-d^2}. $$

Substitute \(d=2\sin(\theta/2)\). Put \(x=\theta/2\):

$$ L=2\cos^{-1}(\sin x)-2\sin x|\cos x|. $$

Hence the expected overlap is

$$ E[L] =\frac1\pi\int_0^\pi \left[2\cos^{-1}(\sin x)-2\sin x|\cos x|\right]dx. $$

Now,

$$ \int_0^\pi \cos^{-1}(\sin x)\,dx =\frac{\pi^2}{4}, $$

and

$$ \int_0^\pi \sin x|\cos x|\,dx=1. $$

Therefore,

$$ E[L] =\frac1\pi\left(\frac{\pi^2}{2}-2\right). $$

So the answer is

$$ \boxed{\frac{\pi}{2}-\frac{2}{\pi}} $$

square units.

Numerically,

$$ \boxed{E[L]\approx0.93418\text{ square units}}. $$

So the expected overlapping area is approximately \(0.934\) square units. For more information you can also check https://wiskundehulpnl.com/

Differential Geometry?