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Point in a circle

A circle of radius 1 is randomly chosen strictly inside a circle of radius 2. A point is chosen, again at random, inside the circle with radius 1. What is the probability this point is closer to the center of the circle with radius 2 than the circumference of the circle with radius 2?

Note by Daniel Liu
4 years ago

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Edited to match the corrected question...

Let \(R\) be the distance of the centre \(O_2\) of the circle \(C\) of radius \(1\) from the centre \(O_1\) of the circle of radius \(2\). Since \(O_2\) can be anywhere inside the circle of radius \(1\) with centre \(O_1\) (shown dotted in the diagram) we have \(\mathbb{P}[R \le r] \,=\, \tfrac{\pi r^2}{\pi} \,=\, r^2\) for any \(0 \le r \le 1\) and hence \(R\) has the probability distribution function \[ f(r) \; = \; \left\{ \begin{array}{ll} 2r & 0 \le r \le 1 \\ 0 & \mbox{o.w.} \end{array} \right. \]

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Given that \(R = r\), the probability that the randomly chosen point in the circle \(C\) is closer to the centre of the original circle than its circumference is the probability that the randomly chosen point lies inside the region common to \(C\) and the (dotted) circle of centre \(O_1\) and radius \(1\), which is \[ \begin{array}{rcl} \mathbb{P}[b > a | R = r] & = & \tfrac{2}{\pi}(\theta - \sin\theta\cos\theta) \\ & = & \tfrac{1}{\pi}\left[2\cos^{-1}(\tfrac{r}{2}) - r\sqrt{1 - \tfrac14r^2}\right] \end{array} \] and hence \[ \begin{array}{rcl} \mathbb{P}[b > a] & = & \displaystyle\int_0^1 \mathbb{P}[b > a|R=r]\,2r\,dr \; = \; \frac{2}{\pi}\int_0^1\left(2r\cos^{-1}(\tfrac{r}{2}) - r^2\sqrt{1 - \tfrac14r^2}\right)\,dr \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac12\pi}^{\frac13\pi}\left(4u\cos u - 4\cos^2u\sin u\right)(-2\sin u)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac13\pi}^{\frac12\pi}\left(4u\sin2u - 2\sin^22u\right)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac23\pi}^\pi \left(v\sin v - \sin^2v\right)\,dv \\ & = & \displaystyle\frac{2}{\pi}\left[ -v\cos v + \sin v - \tfrac12v + \tfrac14\sin2v\right]_{\frac23\pi}^\pi \\ & = & \frac{2}{\pi} \times \frac{1}{8}(4\pi - 3\sqrt{3}) \; = \; 1 - \frac{3\sqrt{3}}{4\pi} \end{array} \]

Mark Hennings - 4 years ago

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Hi,

I tried out a simulation, choosing the coordinates \( (r_1\, \cos \theta+r_2\, \cos \alpha ,r_1\, \sin \theta+r_2\, \sin \alpha )\) at random, where \(0\le r_1, r_2 \le 1\) and \(0\le \theta, \alpha \le 2\,\pi\).

The distance from the center of the bigger circle, \( a=\sqrt{r_1^2+r_2^2+2\, r_1\, r_2\, \cos {(\theta-\alpha)}}\) and \(b = 2 -a\), simulating for 1000000 trials, I got the average \(b > a\) about 0.777, which is more than 0.5865.

Can't see any mistake, though.

Gopinath No - 4 years ago

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Did you assume that \(r_1\) and \(r_2\) were uniform on \([0,1]\)?

Mark Hennings - 4 years ago

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@Mark Hennings Yes, I did..

Gopinath No - 4 years ago

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@Gopinath No The pdf of \(r_1\) and \(r_2\) is \(2x\) for \(0\le x \le 1\). Look at the beginning of my solution.

Mark Hennings - 4 years ago

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@Mark Hennings I saw, but I thought it was required only for deriving the answer analytically.

\(r_1\) is the distance from the center of the circle with radius 2 centered at \((0,0)\), and get the coordinates \( (r_1\cos \theta, r_1 \sin \theta )\), which is the center of the smaller circle. Then select \( (r_2\cos \alpha, r_2 \sin \alpha )\), which is the point within the smaller circle. Then I did my simulation as in the earlier post.

Is there something wrong with this thinking? Aren't the variables independent?

Gopinath No - 4 years ago

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@Gopinath No If you are going to simulate this, you must adjust your simulation so that the chances of choosing \(r_1\) and \(r_2\) correct. You can ignore \(\theta\). Provided that you choose \(r_1\) and \(r_2\) independently from the correct pdf, and choose \(\alpha\) randomly and uniformly in \([0,2\pi)\), all should be well. It will not work if you choose \(r_1\) and \(r_2\) randomly from a uniform distribution.

To model \(r_1\) and \(r_2\) correctly, choose \(r_1^2\) and \(r_2^2\) from a uniform distribution on \([0,1]\). If you do that, your simulation should agree with my theoretical result (my one did).

Mark Hennings - 4 years ago

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@Mark Hennings Oh, I see! It's agreeing now.

Thanks for your help, much appreciated.

Gopinath No - 4 years ago

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Sorry, little typo there. Fixed. That would mean the probability is \(\dfrac{4\pi-3\sqrt{3}}{4\pi}\), correct? I never thought this problem would actually be that hard.

Daniel Liu - 4 years ago

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