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?

  • $\begingroup$ Hint: Distance is the same along circles of center $P$ and radius between $d$ and $d+2r$. $\endgroup$ – Smurf Feb 19 '17 at 15:39
  • $\begingroup$ @Smurf Correct, but what is the pdf and how to calculate the expected value? $\endgroup$ – fhm Feb 19 '17 at 15:44
  • $\begingroup$ 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. $\endgroup$ – Smurf Feb 19 '17 at 15:54
  • $\begingroup$ Did you manage to solve it? $\endgroup$ – Smurf Feb 20 '17 at 18:10
  • $\begingroup$ @Smurf I've been trying to figure out $l(t)$ and integral, yet was unsuccessful. $\endgroup$ – fhm Feb 21 '17 at 6:45

enter image description here

According to this post


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 $$l(R)=2R\arccos\bigl(\frac{R^2-d^2}{2R(d+r)}+\frac{d}{R}\bigl)$$

  • $\begingroup$ 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 solve the integrate over $l(R)$. $\endgroup$ – fhm Feb 21 '17 at 16:17
  • $\begingroup$ Yes, that should do it. Although nothing pretty seems to come out of that. $\endgroup$ – Smurf Feb 21 '17 at 17:20
  • $\begingroup$ 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. $\endgroup$ – Smurf Feb 21 '17 at 17:23
  • $\begingroup$ 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? $\endgroup$ – fhm Feb 26 '17 at 7:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.