To determine the center of the circle given by the equation [tex]\((x-3)^2 + (y-9)^2 = 16\)[/tex], we start by recalling the standard form of a circle's equation:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
Here, [tex]\((h, k)\)[/tex] represents the center of the circle, and [tex]\(r\)[/tex] represents the radius.
Given the equation:
[tex]\[
(x - 3)^2 + (y - 9)^2 = 16
\][/tex]
we can compare this with the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] to identify the values of [tex]\(h\)[/tex] and [tex]\(k\)[/tex].
Upon comparison, it is evident that:
- [tex]\(h = 3\)[/tex]
- [tex]\(k = 9\)[/tex]
Thus, the center of the circle is:
[tex]\[
(h, k) = (3, 9)
\][/tex]
So, the correct answer is:
C. [tex]\((3, 9)\)[/tex]