= Solution
For a radial function, the equation is
$$
n_t=\frac D r\frac{\partial}{\partial r}
\left(rn\frac{\partial n}{\partial r}\right).
$$
Therefore, assuming the population density and radial flux decay at infinity and are regular at the origin,
$$
\frac d{dt}\left(2\pi\int_0^\infty nr\,dr\right)
=2\pi D\left[rn n_r\right]_{0}^{\infty}=0.
$$
Thus $N$ is conserved; it is the total population obtained by integrating density over the plane. This is the global form of the equation's <conservation law>.
Set
$$
n(r,t)=\left(\frac N{Dt}\right)^{1/2}f(x),
\qquad
x=\frac r{(NDt)^{1/4}}.
$$
Then
$$
n_t=\left(\frac N D\right)^{1/2}t^{-3/2}
\left(-\frac12f-\frac14xf'\right),
$$
while the radial diffusion term has the same prefactor and equals
$$
D\nabla\cdot(n\nabla n)
=\left(\frac N D\right)^{1/2}t^{-3/2}
\frac1x\frac d{dx}(xff').
$$
Equating them and multiplying by $x$ gives
$$
\frac d{dx}\left(xff'+\frac14x^2f\right)=0,
$$
so the proposed ansatz is a <similarity solution>.
Regularity at $x=0$ makes the constant of integration zero. Wherever $f>0$,
$$
f'=-\frac x4,
\qquad
f(x)=\frac{x_0^2-x^2}{8}
$$
for some $x_0>0$. The nonnegative weak solution continues as $f=0$ once this parabola reaches zero:
$$
\boxed{f(x)=\left[\frac{x_0^2-x^2}{8}\right]_+.}
$$
This is the <two-dimensional Barenblatt profile for quadratic porous-medium diffusion>; its degenerate diffusivity permits <compact support>.
Indeed, conservation requires
$$
1=2\pi\int_0^\infty f(x)x\,dx
=\frac{\pi x_0^4}{16},
$$
so $x_0=(16/\pi)^{1/4}$ if desired. At time $t$ the populated disk has radius $x_0(NDt)^{1/4}$ and hence area
$$
\boxed{A(t)=\pi x_0^2(NDt)^{1/2}.}
$$
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