To determine the center of the circle represented by the equation \((x + 9)^2 + (y - 6)^2 = 10^2\), we need to compare this equation to the standard form of a circle’s equation, which is:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
Here, \((h, k)\) represents the coordinates of the center of the circle, and \(r\) is the radius.
Given the equation in the problem:
[tex]\[
(x + 9)^2 + (y - 6)^2 = 10^2
\][/tex]
Let's rewrite \((x + 9)^2\) in a form that matches the standard equation format. Notice that:
[tex]\[
(x + 9) = (x - (-9))
\][/tex]
So, \((x + 9)^2\) can be written as:
[tex]\[
(x - (-9))^2
\][/tex]
Similarly, the term \((y - 6)^2\) already matches the standard form.
Now, comparing:
[tex]\[
(x - (-9))^2 + (y - 6)^2 = 10^2
\][/tex]
we see that \((h, k)\) corresponds to \((-9, 6)\).
Therefore, the center of the circle is:
[tex]\[
\boxed{(-9, 6)}
\][/tex]
