What is the average shortest distance between the circle $(x-a)^2+(y-b)^2=r^2$ and a random point lying in it?
This question is just idle curiosity. Basically, it's the same as finding the difference between its radius and the average distance between the random point and its center. Let $D$ denote the shortest distance between the circle $(x-a)^2+(y-b)^2=r^2$ and the random point $P(X,Y)$, then
\begin{equation}
D=r-\sqrt{(X-a)^2+(Y-b)^2}
\end{equation}
We may assume $X$ and $Y$ are independently uniformly distributed in $(0,a)$ and $(0,b)$, respectively. Then its joint pdf is
\begin{equation}
f_{X,Y}(x,y)=f_X(x)\cdot f_Y(y)=\frac{1}{ab}
\end{equation}
Hence the average of $D$ is
\begin{equation}
E[D]=\int_0^b\int_0^a d\ f_{X,Y}(x,y)\ dx\ dy=r-\frac{1}{ab}\int_0^b\int_0^a \sqrt{(x-a)^2+(y-b)^2}\ dx\ dy
\end{equation}
Is my approach correct? If not, how does one find the correct $E[D]$?
Best Answer
If the point lies in a ring radius $x$, thickness $\delta x$, centred at the centre of the circle, then the distance is $r-x$. The ring has area $2\pi x\ \delta x$ and the circle has area $\pi r^2$. So assuming the distribution for the point $P$ is uniform over the disk, the expected distance is $\frac{1}{\pi r^2}\int_0^r(r-x)2\pi x\ dx=\frac{r}{3}$.