 Solution

Conditioned on being inside the circle, the point is a uniform random point in a disk of radius $1/2$. The annulus from $r$ to $r+dr$ has area $2\pi rdr+O(d{r}^2)$. Dividing by the disk's area givesThe radial probability density function is $8r$ on $[0,1/2]$ and integrates to one. Without conditioning, the corresponding annulus probability for a point in the original square is $2\pi rdr+O(d{r}^2)$.