I have a set of N number of points $P=(x,y)$ that are i.i.d uniformly distributed inside the unit circle $C_1$ with center (0,0), each point $P$ represents a point on the circumference of a circle $C_{r_p} = C_{\sqrt{x^2+y^2}}$ where $r_p$ is the radius of the circle that $P$ lies on. Two points are randomly selected from the set, $P_1$ and $P_2$, what is the probability that $P_2$ lies inside the circle $C_{r_1}$ defined by $P_1$. I worked it out to be $\frac{1}{N}\frac{1}{N-1}\frac{1}{3}$ but i'm not this is correct. My workings are: $\Pr(P_1) = 1/N$
$\Pr(P_2) = 1/(N-1)$
For any given point $P$ the probability $Pr(P_2\in C_{r_p}) = A(C_{r_p})/A(C_1)$ where A is the area, and $r_p$ is the radius of the circle defined by $P$
Integrating this over all possible $r_p$'s gives $\frac{1}{A(C_1)}\int^1_0\pi r^2 dr = \frac{1}{\pi}\times\frac{1}{3}\pi = \frac{1}{3}$
Multiply this by the probability of picking the points gives $\frac{1}{N}\frac{1}{N-1}\frac{1}{3}$
Is this correct?
The comments already contain most of what there is to say; I'll write it out as an answer and show why your integration yielded $\frac13$ instead of the correct value of $\frac12$.
First, the probabilities of picking these particular points don't enter into the result. You're not being asked about the probability of picking these points but about the probability that, once you've picked them, the first lies inside the circle defined by the other. As was pointed out in the comments, this has nothing to do with $N$; the entire construction of $2$ points being picked from $N$ points that are uniformly distributed is irrelevant and an unnecessary distraction; at the end of it you simply have two points that are uniformly distributed, and that's all you need to know.
In fact, you don't even need to know that; all you need to know is that they're identically and continuously distributed, since, as achille hui has pointed out the answer $\frac12$ is determined entirely by symmetry, and all you need for this symmetry to obtain is that the distributions are identical and that the radii are almost surely different (which is why the distribution should be continuous).
As to why your integration yielded $\frac13$ instead of $\frac12$: You calculated as if all radii were equally likely. They're not; more of the circle's area is at larger radii than at smaller radii. The density of points at $r$ is proportional to the circumference $2\pi r$ of the circle at $r$, so the probability that you're looking for is
$$ \frac1\pi\frac{\int_0^12\pi r\mathrm dr\pi r^2}{\int_0^12\pi r\mathrm dr}=\frac{\int_0^1r^3\mathrm dr}{\int_0^1r\mathrm dr}=\frac{\frac14}{\frac12}=\frac12\;. $$