To find the center of the circle given by the equation [tex]\((x-3)^2 + (y-9)^2 = 16\)[/tex], we need to recognize that this equation is in the standard form of a circle's equation, which is [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex].
In the standard form equation [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex]:
- [tex]\(h\)[/tex] and [tex]\(k\)[/tex] are the coordinates of the center of the circle [tex]\((h, k)\)[/tex].
- [tex]\(r^2\)[/tex] represents the square of the radius of the circle.
Comparing the given equation [tex]\((x-3)^2 + (y-9)^2 = 16\)[/tex] to the standard form, we can identify the following:
- The term [tex]\((x-3)\)[/tex] corresponds to [tex]\((x-h)\)[/tex], which means [tex]\(h = 3\)[/tex].
- The term [tex]\((y-9)\)[/tex] corresponds to [tex]\((y-k)\)[/tex], which means [tex]\(k = 9\)[/tex].
- The term [tex]\(16\)[/tex] corresponds to [tex]\(r^2\)[/tex], which means [tex]\(r = \sqrt{16} = 4\)[/tex]. However, the radius is not needed to determine the center.
Thus, the center of the circle [tex]\((h, k)\)[/tex] is [tex]\((3, 9)\)[/tex].
The correct choice is:
D. [tex]\((3, 9)\)[/tex]
