Average distance from circle's center to a point
What is the average distance from a point inside a circle to the circle's center?
I came across this problem, and I've heard the solution is $\frac23R$ for the radius $R$.
So, I tried to tackle this myself:
- Let there be two concentric circles $C_1$ and $C_2$ with radii $R$ and $r$ ($R\gt r$).
- We pick a random point $P$ inside $C_1$. If there's a $50\%$ chance that $P$ is inside $C_2$, then $r$ is the average distance from $P$ to the center.
Then: $$[C_1]=\frac12[C_2]$$ $$\not\pi r^2=\frac12\not\pi R^2\implies r=\frac12\sqrt2\cdot R$$
...which doesn't match the answer of $\frac23R$ (though, I should say $\frac12\sqrt2\approx0.707106781187$ which is quite close to $\frac23$).
Why is my reasoning wrong? If it's because $r$ isn't the average distance, why's that? I can't really make sense of this.
Thanks!
1 answer
The following users marked this post as Works for me:
| User | Comment | Date |
|---|---|---|
| TheCodidacter, or rather ACodidacter | (no comment) | Jun 14, 2023 at 10:38 |
The problem is indeed that you aren't computing the average distance. What you're computing is a median distance. Essentially, you are equally weighting points that are just slightly outside $C_2$ and points that are near the boundary of $C_1$, but those contribute different amounts to the average distance. To reinforce this, note that if you did an arbitrary area-preserving transformation that maintained which points were in or out of $C_2$, then you could have half the points be at any arbitrary distance allowing you to increase the average distance arbitrarily. Imagine $C_2$ being the cross-section of a metal cylinder and everything outside of $C_2$ (and in $C_1)$ being goop that you could stretch arbitrarily. You could stretch that goop into a dumbbell shape with a very long and narrow connecting part.
A typical, fairly generic definition of the average value, $\bar f$, of a function, $f$, over a continuous domain, $D$, is $\bar f = \int_D f(x) dx / \int_D dx$. In probabilistic language, this would be the expected value of $f$ if $D$ were the sample space with uniform probability for each point. In this case, $D$ is the disc of radius $R$, so the denominator integral is just its area, namely $\pi R^2$. The numerator integral can be written as $\int_0^{2\pi}\int_0^R r^2 drd\theta$. The perhaps not immediately obvious part is why the integrand is $r^2$ and not just $r$. Intuitively, this is because changing $\theta$ by a small amount changes where you are in direct proportion to the distance from the center of rotation, i.e. $r$. If the $r^2$ isn't obvious to you, you should try to derive it yourself. One, somewhat tedious, approach is to rewrite the integral in Cartesian coordinates and do a change of variables. A more direct way would be to look up what polar coordinates means not just for the coordinates themselves but also the tangent vectors.
The result you quoted then follows through basic integration, $$\int_0^{2\pi}\int_0^R r^2 dr d\theta = 2\pi\int_0^R r^2 dr = 2\pi R^3/3$$ which is divided by $\pi R^2$.
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