What is the center of the circle given by the equation [tex]$(x-3)^2+(y-9)^2=16$[/tex]?

A. [tex]$(-9,-3)$[/tex]
B. [tex]$(-3,-9)$[/tex]
C. [tex]$(3,9)$[/tex]
D. [tex]$(9,3)$[/tex]



Answer :

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]