The problem provides the equation of a circle, $(x-2)^2 + (y+3)^2 = 16$. The task is to find the center and the radius of the circle.

1. Problem Description

The problem provides the equation of a circle, $(x-2)^2 + (y+3)^2 = 16$. The task is to find the center and the 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 the given equation $(x-2)^2 + (y+3)^2 = 16$ with the general equation, we can identify the center and radius.

We have:

$x-h = x-2$, so $h = 2$.

$y-k = y+3$, so $-k = 3$, which means $k = -3$.

$r^2 = 16$, so $r = \sqrt{16} = 4$.

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

3. Final Answer

Center: $(2, -3)$

Radius: $4$