mean distance from a point in a circle to its boundary ( circumference)?

Question

I have perused the solution to the average distance from a point in a ball to a point on its boundary. I don't quite understand it. However, it seems likely that the analogous problem 'Average distance from a point in a circle to the circumference (boundary)' may be easier to solve. This is the problem that I want to solve.

Answer 1

For the unit circle, the distance from a point inside to the circumference is $1-r.$ If this is integrated using polar coordinates to get the average distance, one would integrate $(1-r)rdr\ d\theta$ over the interior of the circle, then divide by the area of the circle to get the average distance.

Answer 2

First, let's make clear what am I solving.

If we have a circle with radius $R$ then the center is at a distance $R$ from any point on the circumference, therefore, the average distance of the center to the circumference is also $R$.

If we take a point $P$ other than the center then it will be closer to some points on the circumference and further from others than it would be if $P$ where on the center of the circle. I understand that the question asks what would be the average distance from this point $P$ to all the points on the circumference.

Assuming that I've understood correctly, then matt biesecker's answer is correct and I'll elaborate on why.

First, let $P$ have coordinates $(h, 0)$. If $P$ didn't have those coordinates we could simply rotate the circle to make it have that position.

Now, a point on the circumference will have coordinates $(x, y)$.

We know that the distance between two points is given by $d = \sqrt{(x_2-x_1)^2 + (y_2 - y_1)^2}$.

Substituing the coordinates of our point inside the circle and our arbitray point on the circumference we get the following: \begin{equation} d = \sqrt{(x-h)^2 + y^2} \end{equation}

If we use polar coordinates this will become way easier, let's recall that, to convert from Cartesian to polar coordinates we have:

$x = r\,Cos(\theta)$

$y = r\, Sin(\theta)$

In this particular case, because we want all the points to be on the circumference with radius $R$ we get:

$x = R\,Cos(\theta)$

$y = R\, Sin(\theta)$

Pluging this in our expression for distance we get:

\begin{equation} d(\theta) = \sqrt{(R\,Cos(\theta)-h)^2 + (R\,Sin(\theta))^2} \end{equation}

\begin{equation} d(\theta) = \sqrt{(R\,Cos(\theta)-h)^2 + R^2 \, Sin^2(\theta)} \end{equation}

This last expression lets us find the distance between our point $P$ and any point on the circumference with radius $R$ at an angle $\theta$.

But we don't want that, do we? We whant the average distance from $P$ to all of the points on the circumference. Let's recall how to get the average value of a function:

\begin{equation} \bar{f(x)} = \frac{1}{b-a}\int_{a}^{b} f(x) \end{equation}

In our problem, $\theta$ goes from $0$ to $2\pi$, in this case we get:

\begin{equation} \bar{d} = \frac{1}{2\pi}\int_{0}^{2\pi} \sqrt{(R\,Cos(\theta)-h)^2 + R^2 \, Sin^2(\theta)} \, d\theta \end{equation}

Which is equivalent to matt biesecker's answer.

Answer 3

Building from Bernardo Morales' foundation:

Consider a Circle with radius $R$ centered at the origin of a polar coordinate plane.
Let $P$ be a point such that $P = (h,0)$. $P$ is the point from which we will measure the average distance to the circumference.

Represent points in the circle as $Q = (R,\theta)$ such that $\theta \in [0, \pi]$ and label the distance between $P$ and $Q$ as $D$. From the law of cosines, we get:

$$ D^2=R^2+h^2-2Rh\cos\theta $$ $$ D = \sqrt{R^2+h^2-2Rh\cos\theta} $$

Since $D$ is a function of $\theta$, we can find the average value of $D$ as $\theta$ goes from $0$ to $\pi$:

$$ \bar{D}=\frac{1}{\pi}\int_0^\pi D(\theta) d\theta $$ $$ \bar{D}=\frac{1}{\pi}\int_0^\pi \sqrt{R^2+h^2-2Rh\cos\theta}\cdot d\theta $$

Here, $\theta$ ranges only from $0$ to $\pi$ because of the symmetry of the circle about the line $\theta=0$. This is equivalent to the solutions provided by Bernardo Morales and matt biesecker.