Identify the vertices and center of the ellipse. $x^2+36y^2-8x-72y+16=0$

A) vertices: $(-5,-1),(13,-1)$ center: (-4,-1)
B) vertices: $(3,1),(5,1)$ center: (4,1)
C) vertices: $(-4,-7),(-4,5)$ center: $(-4,-1)$
D) vertices: $(-2,1),(10,1)$ center: $(4,1)$
E) vertices: $(-4,-2),(-4,4)$ center: $(-4,-1)$

To identify the vertices and center of the ellipse, we need to divide both sides of the equation by 16 to get it in standard form: $\frac{(x-1)^2}{16}+\frac{(y-1)^2}{4}=1$. The center is given by $(h,k)$, so the center is $(1,1)$. The vertices lie on the major axis, which has length $2a$. In this case, $2a = 4\sqrt{2}$, so $a = 2\sqrt{2}$. Since the center is $(1,1)$, the vertices lie at $(1+2\sqrt{2},1)$ and $(1-2\sqrt{2},1)$. These points match with the vertices given in answer D.