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:
$$F_R(r) = \mathbb{P}(R \leqslant r) = \frac{A(r)}{A(1)} = \frac{\pi r^2}{\pi} = r^2
\quad \quad \quad 0 \leqslant r \leqslant 1.$$
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 $R$ from the origin to the point does not have a uniform distribution; as expected, it is more likely to be nearer to the outer edge of the circle than its origin.