The problem gives the equation of a circle as $(x-2)^2 + (y+3)^2 = 16$. We need to find the center and radius of the circle.

1. Problem Description

The problem gives the equation of a circle as $(x-2)^2 + (y+3)^2 = 16$. We need to find the center and radius of the circle.

2. Solution Steps

The general equation of a circle with center $(h, k)$ and radius $r$ is given by:

$(x-h)^2 + (y-k)^2 = r^2$

Comparing this with the given equation $(x-2)^2 + (y+3)^2 = 16$, we can identify the center and the radius.

From the equation, we have:

$h = 2$

$k = -3$ (since $y+3 = y - (-3)$)

$r^2 = 16$

Therefore, $r = \sqrt{16} = 4$

3. Final Answer

The center of the circle is $(2, -3)$ and the radius is $4$.