Write the ellipse equation in standard form given its foci are at (2,0) and (-2,0) and its latus rectum is 6.

The foci are $F_1(2,0)$ and $F_2(-2,0)$.
The center of the ellipse is the midpoint of the foci:

The distance from the center to each focus is $c$.

Since the foci are on the x-axis, the major axis is horizontal. The standard form of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The latus rectum $L = 6$. The formula for the latus rectum is $L = \frac{2b^2}{a}$.

For an ellipse, $c^2 = a^2 - b^2$.
Substitute $c=2$:

Substitute equation (1) into equation (2):

Rearrange into a quadratic equation:

Factor the quadratic equation:

Since $a$ must be positive, $a=4$.
Now find $b^2$ using equation (1):

And $a^2 = 4^2 = 16$.

Substitute $a^2 = 16$ and $b^2 = 12$ into the standard form:

The equation of the ellipse in standard form is $\boxed{\frac{x^2}{16} + \frac{y^2}{12} = 1}$.

For the $x$ terms: Take half of the coefficient of $x$ ($-\frac{2}{3}$), which is $-\frac{1}{3}$, and square it: $\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$.
For the $y$ terms: Take half of the coefficient of $y$ ($-1$), which is $-\frac{1}{2}$, and square it: $\left(-\frac{1}{2}\right)^2 = \frac{1}{4}$.

The center $(h,k)$ is $\left(\frac{1}{3}, \frac{1}{2}\right)$.
The radius squared $r^2 = 1$, so the radius $r = \sqrt{1} = 1$.

The center of the circle is $\boxed{\left(\frac{1}{3}, \frac{1}{2}\right)}$ and the radius is $\boxed{1}$.