Let $A$ denote, in the plane, the area enclosed by the perimeter $P=\partial A$, and $K=\frac{1}{2\pi} \int_0^{2\pi}\kappa(\theta)d\theta$ the mean curvature of $P$ (i.e. the average of the local curvature $\kappa$ all along $P$). We adopt the polar coordinates $(r,\theta)$. We know that when $A$ reduces to a disk (or $P$ reduces to a circle) of radius $R$, then $K=1/R$.
I am trying to prove (or disprove) the following result:
$K=1/\bar{R}$,
where $\bar{R}$ is the radius of the equivalent disk $\bar{A}$, that is, the disk with same area as $A$.
My starting point was to compute $\kappa(\theta) = \frac{r^2+2r'^2-r r''}{(r^2+r'^2)^{3/2}}$, where the prime denotes the derivative of $r(\theta)$. We know that $A=\bar A$, i.e., $\int\int_A rdrd\theta=\pi \bar R^2$, by definition of the equivalent disk (I abusively note $A$ the area of $A$).
Then, upon applying Green's theorem (if not mistaken), $K=\frac{1}{2\pi} \int \int_A \frac{\partial\kappa}{\partial r} dr d\theta$. From there, I would combine the two previous equations, but calculations become rather tedious, and I'm not sure if it leads anywhere. If this result is true, there exists maybe a more straightforward way of proving/disproving it?
Any help appreciated, thanks!