Let $u(r,t)$ solve the heat equation in polar coordinates in the complement of the disk:
$$ \partial_t u(r,t)= \frac{1}{2}\frac{1}{r}\partial_r r \partial_r u(r,t), \quad r>1$$
with initial condition $u(r,0)=0$ and boundary condition $u(1,t)=1$.  I claim that the number you seek is 
$$ A_t = \pi + 2\pi\int_1^\infty u(r,t)r dr.$$
(This is because $u(r,t)$ gives the probability that a Brownian particle released at radius $r$ has hit the disk by time $t$. The first term $\pi$ is just the area of the disk at time $0$. I have transferred the problem to one in which your particle remains fixed and potential target points move.)  

There is an exact expression for $u$ (see [Wendel, J. G. "Hitting spheres with Brownian motion". Ann. Prob. 8, 164 (1980).][1], which I learned of from the answer to this [mathoverflow question][2], but ultimately traced back to a [paper of Carslaw and Jaeger][3] from 1940).  It is most easily expressed in terms of the Laplace transform $F(r,\lambda)=\int_0^\infty u(r,t) e^{-\lambda t}$:
$$F(r,\lambda) = \frac{K_0(\sqrt{2\lambda}r)}{\lambda K_0(\sqrt{2\lambda})}, \quad r\ge 1,$$
where $K_0$ is a modified Bessel function of the second kind and order $0$. In your case the integral over $r$ can be done exactly since $\int r K_0(r) dr = - K_1(r)$. So 
$$ \int_0^\infty A_t e^{-\lambda t} d t \ = \ \frac{\pi}{\lambda} + \frac{2\pi K_1(\sqrt{2\lambda})}{\lambda (2\lambda)^{\frac{3}{2}}K_0(\sqrt{2\lambda})}.$$
You can recover $A_t$ by analytic continuation of the right hand side to complex $\lambda$ and Fourier inversion.  


  [1]: http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aop/1176994833
  [2]: http://mathoverflow.net/questions/96703/finite-time-hitting-probabilities-for-brownian-motion-in-the-plane
  [3]: http://plms.oxfordjournals.org/content/s2-46/1/361.full.pdf