The equation of the circle in polar coordinates is $\rho = 2\cos\theta$.

1. Convert the polar equation to Cartesian coordinates:
To find the center of the circle in Cartesian coordinates, we start by expressing $\rho$ and $\theta$ in terms of $x$ and $y$. For a point in polar coordinates $(\rho, \theta)$, the corresponding Cartesian coordinates are given by $x = \rho\cos\theta$ and $y = \rho\sin\theta$. Substituting $\rho = 2\cos\theta$ into these equations eliminates $\rho$:

$x = 2\cos^2\theta$ and $y = 2\cos\theta\sin\theta$.

Using the double-angle identity $\cos(2\theta) = 2\cos^2\theta - 1$, we can solve for $\cos^2\theta$:

$\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$.

Thus, $x = 1 + \cos(2\theta)$.

We apply the identity $\sin(2\theta) = 2\sin\theta\cos\theta$ to express $y$:

$y = \sin(2\theta)$.

Now we are able to express $x$ and $y$ without explicit dependence on $\theta$:

$(x - 1)^2 + y^2 = 1$.

This is the Cartesian equation of a circle with the center at $(1, 0)$ and radius $1$.

2. Calculate the distance from the point to the circle's center:
The given point $P(2, \frac{\pi}{3})$ in Cartesian coordinates is $(1, \sqrt{3})$ since $x = \rho\cos\theta = 2\cos\frac{\pi}{3} = 2 \cdot \frac{1}{2} = 1$ and $y = \rho\sin\theta = 2\sin\frac{\pi}{3} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$.

The distance $d$ from point $P(1, \sqrt{3})$ to the circle's center at $(1, 0)$ is:

$$d = \sqrt{(1 - 1)^2 + (\sqrt{3} - 0)^2} = \sqrt{0 + 3} = \sqrt{3}.$$

So the distance is $\boxed{\sqrt{3}}$.