In a unit circle what is the average distance to the center of circle? In a circle with radius 1 what is the average distance between a randomly placed point and the center of the circle?
So far I have tried a few things but gotten different results.
Approach 1:
My idea here is to take the weighted average of all distances where the circumference of the corresponding circle is the weight. The definition of the weighted average is:
$\frac{\sum_{0} ^{n} {w_n v_n}}{\sum_{0} ^{n} {w_n}}$
where $w_n$ and $v_n$ are lists of weights and values
This leads me to:
$\frac{\int_0^1 {2\pi r^2}dr}{\int_0^1 {2\pi r} dr} = \frac{2}{3}$
I believe this is correct due to a computer simulation I made, but I could have done that wrong.
Approach 2: Since the probability of a point falling in a certain section of a circle is proportional to the area of the section my other idea was to find the distance where the probability of falling inside and outside is equal, or the area of the inner circle is equal to the area of the rest of the circle.
$\pi x^2 = \pi 1^2 - \pi x^2$
$x = \frac{\sqrt{2}}{2}$
Why is this different than my first answer? Which approach is wrong and why?
 A: Let $X$ denote the random variable that represents the distance of the random chosen point from the centre of the circle. Let $f$ and $F$ denote the pdf and cdf of $X$. Note that $\Pr(X \leq x)$ is simply the ratio of the area of disc of radius $x$ to that of the entire disc. Thus,
$$ F(x) = \Pr(X \leq x) = \frac{\pi x^2}{\pi \cdot 1^2} = x^2.$$
Consequently, $f(x) = 2x$. The required value is
$$ \mathbb{E}[X] = \int_{0}^{1} x f(x) \ \mathrm{d}x = \int_0^1 2x^2 \ \mathrm{d}x = \frac{2}{3}.$$
As you suggested, your first answer is correct. In the second approach, you ended up finding the median of $X$ instead of mean. Mean does not mean the probability mass should be equal, it just represents the weighted average of all possible values.
A: The question is ambiguous, and the answer depends on the algorithm used to randomly select a point.
For example, suppose that you randomly select an angle $\theta$ such rhat $0^\circ \leq \theta < 360^\circ.$
Then, suppose that you randomly select a point on the line segment between the center of the circle and the circle boundary, where  the line segment makes an angle $\theta$ with the horizontal.
The above algorithm, which is a perfectly valid way of selecting a random point, will result in the expected distance being $~\dfrac{r}{2},~$ where $r$ is the radius of the circle.
A: Your first method is correct. The average distance is just the integral of the distance over the circle divided by its area. Given a circle of radius 1, the integral of the distance is
$$
\int\int_{〇} r \;dx\;dy=\int_0^1\int_0^{2\pi}r^2drd\theta={2\pi\over3}
$$
and the area is $\pi$, so the average distance of a random point is 2/3. If the circle has radius $R$ then just multiply by $R$, so $\frac23R$.
In the second method, you find two shapes which have equal areas, and the probability of the point falling in the two areas is indeed equal, but you haven't multiplied that by the distance, but just used the radius of the inner circle to be the distance. But the radius of the inner circle is not related to the average distance in any simple way.
To see that another way, imagine cutting the circle into two semicircles. The chance of the point falling into either semicircle is the same, but you can't get the average distance from the origin like this.
A: Your second approach is false for what you wanted to achieve (if I understood you correctly), Suzu finished your second approach.
For why you may get different results you may wan to check this link.
A: The question cannot be answered since the problem is not well posed without specifying what exactly it means to "randomly place" a point. See the analogous problem know as the Bertrand paradox.
