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


Sagot :

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]