Let $C_1$ be our $\color{blue}{\text{blue circle}}$, and $C_2$ our $\color{darkorange}{\text{orange circle}}$ with the following coordinates. Note that $C_2$ lies at $(0,R)$ which is an important detail for this problem.

Our two circles' equations are
$\begin{eqnarray}
C_1&:& x^2+y^2 &= R^2 \tag{1}\\
C_2&:& (x-R)^2+y^2 &= r^2 \tag{2}
\end{eqnarray}$
Combining $(1)$ and $(2)$ and solving for $x$ we get
$$(x-R)^2+(R^2-x^2) = r^2 \\ x_1 = \dfrac{2R^2-r^2}{2R}$$
So what did we just find? We found the $x$-coordinate at which the two circles intersect.
Using the formula to find an area for a circular segment. Let $R'$ be a radius, and $d$ a distance from a circle's centre to a chord. For our case, our chord would be a vertical line going through the intersection points of our circles.
$$A(R',d) = R'^2\cdot\cos^{-1}\left(\frac{d}{R'}\right) - d\sqrt{R'^2-d^2}$$
Using this, and plugging in respectively value for our circles we get the following
$$A(R,x_1) = R^2\cdot\cos^{-1}\left(\frac{x_1}{R}\right) - x_1\sqrt{R^2-x_1^2}$$
So where are we now? We've calculated the shaded area below, and we need to find $d$ so that we can do the same for the other circular segment.

Well, we know that the distance between our circles is $R$. So why not put $d = R-x_1$. Hence,
$$A(r,d) = r^2\cdot\cos^{-1}\left(\frac{d}{r}\right) - d\sqrt{r^2-d^2}$$
This would mean our $A_{total} = A(R,x_1) + A(r,d)$. Writing this out becomes tedious, but using Mathematica gives,
$$A_{total} = r^2\cos^{-1}\left(\frac{r^2}{2R\cdot r}\right) + R^2\cos^{-1}\left(\frac{2R^2-r^2}{2R^2}\right) - \frac{1}{2}\sqrt{r^2\cdot 2R(2R+r)}$$
Now, using our $A_{total}$ we can find values for $r$ by specifying $R$ and the area ratio you want. I.e.
If $R = 2$ and you want the area between the circles to equal half of $C_1$'s area, you would put:
$$R = 2: A_{total} = \frac{1}{2}\pi\cdot R^2 = 2\pi$$
Let Mathematica calculate $A_{total}$ numerically after specifying $R$, and you will find it gives you two solutions. Use the positive numerical solution and that's the radius you were looking for.
NB! By trying different values for $R$ and dividing the result $r$ with $R$, respectively, you'll find $\frac{R}{r} \approx 1.15872847 \ldots$. (OEIS A133731)
And as such, for this particular problem, the radius $r$ is always near $1.16\cdot R$ length units.