You are extremely close to the answer.  You have already noted that the area of a circle with radius $r$ is $A(r) = \pi r^2$.  Since you are selecting a uniform point on the circle, this means that you have the cumulative distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2.$$

The reasoning behind this is that with a uniform distribution we have a flat density over the circle.  So the probability of falling within a given area is the relative size of that area compared to the size of the circle.  This gives the corresponding density function:

$$f_R(r) = 2r \quad \quad \text{for all } 0 \leqslant r \leqslant 1.$$

As you can see from this result, although the chosen point is uniformly distributed on the circle, the resulting distance $R$ to the point does not have a uniform distribution; as expected, it is more likely to bear nearer to one than nearer to zero.