×

# 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

Sort by:

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 $f(r) \; = \; \left\{ \begin{array}{ll} 2r & 0 \le r \le 1 \\ 0 & \mbox{o.w.} \end{array} \right.$

alternatetext

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 $\begin{array}{rcl} \mathbb{P}[b > a] & = & \displaystyle\int_0^1 \mathbb{P}[b > a|R=r]\,2r\,dr \; = \; \frac{2}{\pi}\int_0^1\left(2r\cos^{-1}(\tfrac{r}{2}) - r^2\sqrt{1 - \tfrac14r^2}\right)\,dr \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac12\pi}^{\frac13\pi}\left(4u\cos u - 4\cos^2u\sin u\right)(-2\sin u)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac13\pi}^{\frac12\pi}\left(4u\sin2u - 2\sin^22u\right)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac23\pi}^\pi \left(v\sin v - \sin^2v\right)\,dv \\ & = & \displaystyle\frac{2}{\pi}\left[ -v\cos v + \sin v - \tfrac12v + \tfrac14\sin2v\right]_{\frac23\pi}^\pi \\ & = & \frac{2}{\pi} \times \frac{1}{8}(4\pi - 3\sqrt{3}) \; = \; 1 - \frac{3\sqrt{3}}{4\pi} \end{array}$

- 4 years ago

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.

- 4 years ago

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

- 4 years ago

Yes, I did..

- 4 years ago

The pdf of $$r_1$$ and $$r_2$$ is $$2x$$ for $$0\le x \le 1$$. Look at the beginning of my solution.

- 4 years ago

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?

- 4 years ago

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).

- 4 years ago

Oh, I see! It's agreeing now.

Thanks for your help, much appreciated.

- 4 years ago

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.

- 4 years ago