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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 given by the equation [tex]\( (x+9)^2 + (y-6)^2 = 10^2 \)[/tex], we need to compare this equation to the standard form of the equation of a circle.

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

By comparing the given equation [tex]\( (x+9)^2 + (y-6)^2 = 10^2 \)[/tex] with the standard form, we can identify the coordinates of the center [tex]\((h, k)\)[/tex].

1. The term [tex]\( (x+9)^2 \)[/tex] implies that [tex]\( h = -9 \)[/tex]. This is because [tex]\( (x - (-9))^2 = (x + 9)^2 \)[/tex].

2. The term [tex]\( (y-6)^2 \)[/tex] implies that [tex]\( k = 6 \)[/tex]. This is because [tex]\( (y - 6)^2 \)[/tex] is already in the correct format.

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

Given the options:
- [tex]\((-9, 6)\)[/tex]
- [tex]\((-6, 9)\)[/tex]
- [tex]\((6, -9)\)[/tex]
- [tex]\((9, -6)\)[/tex]

The correct answer is:
[tex]\[ (-9, 6) \][/tex]

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