We are given the equation of a circle $(x+3)^2 + y^2 = 49$ and asked to find the coordinates of the center of the circle.

1. Problem Description

We are given the equation of a circle $(x+3)^2 + y^2 = 49$ and asked to find the coordinates of the center of the circle.

2. Solution Steps

The standard 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 $(x+3)^2 + y^2 = 49$ with the standard equation, we can rewrite the given equation as:

$(x - (-3))^2 + (y - 0)^2 = 7^2$

From this, we can identify the center as $(h, k) = (-3, 0)$.

3. Final Answer

The center of the circle is $(-3, 0)$.