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 find the center of the circle represented by the given equation [tex]\((x+9)^2+(y-6)^2=10^2\)[/tex], we need to understand the general form of the equation of a circle.

The general form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle, and [tex]\(r\)[/tex] is the radius.

Let's compare the given equation [tex]\((x + 9)^2 + (y - 6)^2 = 10^2\)[/tex] with the general form.

1. [tex]\((x + 9)^2\)[/tex] can be written as [tex]\((x - (-9))^2\)[/tex].
2. [tex]\((y - 6)^2\)[/tex] is already in the correct form, [tex]\((y - 6)^2\)[/tex].

From this comparison:
- The term [tex]\((x - h)\)[/tex] becomes [tex]\((x - (-9))\)[/tex], so we find [tex]\(h = -9\)[/tex].
- The term [tex]\((y - k)\)[/tex] is already [tex]\((y - 6)\)[/tex], so we find [tex]\(k = 6\)[/tex].

Therefore, the center of the circle [tex]\((h, k)\)[/tex] is [tex]\((-9, 6)\)[/tex].

So, the correct answer is:

[tex]\[ (-9, 6) \][/tex]

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