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Point $P$ is located at distance $d$ from a circle with radius $r$ (that is $d+r$ from the center of circle). What would be the expected value of the distance between the point $P$ and any random (uniform) point in the circle and why?

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    Hint: Distance is the same along circles of center $P$ and radius between $d$ and $d+2r$. – 2017-02-19
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    @Smurf Correct, but what is the pdf and how to calculate the expected value? – 2017-02-19
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    Roughly speaking, the pdf should be like this: $$f(t)=t\cdot l(t)$$ where $$l:t\mapsto \text{measure of the set of points at length $t$ from $P$}$$ So the problem would be to explicitly calculate the function $l$, here you can use my previous hint. – 2017-02-19
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    Did you manage to solve it? – 2017-02-20
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    @Smurf I've been trying to figure out $l(t)$ and integral, yet was unsuccessful. – 2017-02-21
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    I think I managed to obtain $l(t)$, but I am not able to integrate, I'll write down what I have done so far later. Where did you find this problem? – 2017-02-21
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    @Smurf Thank you, it would be helpful. I modeled this problem based on a scenario in wireless networks. Stochastic geometry is used to model a couple of problems there. – 2017-02-21
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    I also got a hint from a friend (Jan) which says the $pdf$ over which I need to integrate $t$ might be the arc-length of intersection of the circle with the circle with radius $t\in[d,d+2r]$ centered at $P$. Seems close to your approach. – 2017-02-21
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    Yes, that is exactly my approach, although as you can see is extremely messy, probably I missed something and can be simplified. – 2017-02-21

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enter image description here

According to this post

https://en.wikipedia.org/wiki/Circular_segment

we need to calculate $d+x$ (notice that our $d$ is not equal to the one in the post) in order to get $\theta$ and finally $s$.

Let's start by calculating $x$, in the picture there are two right triangles that share a side, let's call it $y$, then

$$\left\{\begin{matrix}R^2&=&(d+x)^2+y^2\\r^2&=&(r-x)^2+y^2\end{matrix}\right.\Rightarrow R^2-(d+x)^2=r^2-(r-x)^2\Rightarrow x=\frac{R^2-d^2}{2(d+r)}$$

thus $$\theta=2\arccos\bigl(\frac{d+x}{R}\bigl)=2\arccos\bigl(\frac{R^2-d^2}{2R(d+r)}+\frac{d}{R}\bigl)$$ and finally $$l(R)=2R\arccos\bigl(\frac{R^2-d^2}{2R(d+r)}+\frac{d}{R}\bigl)$$

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    I appreciate your effort. You actually conclude with $l(R)=R\theta$ where $R\in[d,d+2r]$ is chosen randomly ($R$ instead of $t$?). May [Integration by parts](https://www.cymath.com/answer.php?q=int(arccos(y)%2Cy)) solve the integrate over $l(R)$. – 2017-02-21
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    Yes, that should do it. Although nothing pretty seems to come out of that. – 2017-02-21
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    Another way might be using coordinates (cartesian or polar) and do a double integral. At first I thought this way would be more messy, until I saw the integral we got here. – 2017-02-21
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    In Matlab: let's set `d=30`, `r=10`. Define `f` as `f=@(x)2.*x.*acos(((x.^2-d^2)./(2.*x.*(r+d)))+d./x).*(1/((d+2‌​*r)-d));` (Uniform pdf at the end). Now the integration `integral(f,d,d+(2*r))` returns the same value as if `d=60`, `r=10` or `d=200`, `r=10`. The solution only depends on r. Why? – 2017-02-26