= 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 $2\pi r\,dr+O(dr^2)$. Dividing by the disk's area gives
$$
\boxed{P(r<R<r+dr\mid R<1/2)=8r\,dr+O(dr^2),\qquad0\leq r<1/2.}
$$
The 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 $2\pi r\,dr+O(dr^2)$.
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