Point in a circle - Distance from points at the circumference? Let's say we have a random point inside a circle . My intuition tells me that that If I draw vectors from all the circumference points to that point and find their sum the result will be zero.
Am I correct ? How can we prove this ?
Edit: I am asking about the sum of the 'vectors'
 A: What is the sum of the vectors? There are infinitely many points on the circumference. 
If we can define the sum clearly, we may attack the problem by the decomposition of vectors.
Let the random point be $A$, the centre of the circle be $O$ and an arbitrary point on the circumference be $P$.
Then $\overrightarrow{AP}=\overrightarrow{OP}-\overrightarrow{OA}$.
$$\sum_P \overrightarrow{AP}=\sum_{P}\overrightarrow{OP}-\sum_P\overrightarrow{OA}$$
The term $\displaystyle \sum_P\overrightarrow{OA}$ will cause troubles.
I think it is better to consider the averagew of the vectors.
If we can define the 'average' clearly and let the average of $\displaystyle \sum_P \overrightarrow{AP}$ be $u$ and the average of $\displaystyle \sum_P \overrightarrow{OP}$ be $v$, then we have
$$u=v-\overrightarrow{OA}$$
In the heading, OP raise the problem on the distance. My approach is on the vectors. Distances will be another story.
A: Take the special case when when random point is at center. May be you are implying angular separation $2 \pi/n$ when $n$ is large. Clearly your intuition is incorrect.
If you however want their vector sum then for this special case indeed it is zero, because an Engineering drawing results in a vector polygon diagram as a closed circle.
And if there is an eccentricity $e$ then such a vector results in $ n\cdot e$ because each eccentric vector is sum of vector $e$ and (total sum zero) radial vectors.
