E.g. in Rudins real and complex analysis, chapter 11, you'll find the following statement:
If $u$ is a continuous and real valued function on the boundary of the disc $D(a; R)$ and if $u$ is defined in $D(a; R)$ by the Poisson integral
$u(a+r e^{i\theta}) = \frac{1}{2\pi} \int\limits_{-\pi}^{\pi} \frac{R^2-r^2}{R^2+r^2-2 rR\cos(\theta-t)} u(a+Re^{it}) dt $
then u is continuous on the closeure of $D(a;R)$ and harmonic in $D(a,R)$.
Your integral is -- up to a factor $\frac{R^2- r^2}{2\pi}$ -- the specialization of that integral to the constant function $u=1$. If you introduce this factor you get the harmonic extension of the constant function (which has to be constant by the maximum principle) $1$ at the origin. Hence your integral is the inverse of that factor. (You need to rename to your $r, \rho$, of course)
(Admittedly, you probably have to evaluate that integral directly to prove the aforementioned statement :-))