What is the center of a circle represented by the equation [tex](x+9)^2+(y-6)^2=10^2[/tex]?

A. [tex]\((-9, 6)\)[/tex]
B. [tex]\((-6, 9)\)[/tex]
C. [tex]\((6, -9)\)[/tex]
D. [tex]\((9, -6)\)[/tex]



Answer :

To determine the center of a circle represented by the equation [tex]\((x + 9)^2 + (y - 6)^2 = 10^2\)[/tex], we need to compare it to the standard form of a circle's equation, which is:

[tex]\[ (x - h)^2 + (y - k)^2 = r^2. \][/tex]

In this standard form:
- [tex]\((h, k)\)[/tex] is the center of the circle.
- [tex]\(r\)[/tex] is the radius of the circle.

Given the equation [tex]\((x + 9)^2 + (y - 6)^2 = 10^2\)[/tex], we observe the following:

- The term [tex]\((x + 9)\)[/tex] can be rewritten in the form [tex]\((x - (-9))\)[/tex]. This translation means that [tex]\(h = -9\)[/tex].

- The term [tex]\((y - 6)\)[/tex] already matches the form [tex]\((y - k)\)[/tex], indicating that [tex]\(k = 6\)[/tex].

Therefore, the center of the circle is [tex]\((h, k)\)[/tex], which translates to:

[tex]\[ (h, k) = (-9, 6). \][/tex]

So, the correct answer is [tex]\((-9, 6)\)[/tex].