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To identify the center of the circle given by the equation [tex]\((x+9)^2 + (y-6)^2 = 10^2\)[/tex], we need to understand the standard form of a circle's equation. A circle's equation in standard form 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 of the circle.
Comparing the given equation [tex]\((x+9)^2 + (y-6)^2 = 10^2\)[/tex] with the standard form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], we can determine the values of [tex]\(h\)[/tex] and [tex]\(k\)[/tex].
1. The term [tex]\((x-h)^2\)[/tex] in the standard form corresponds to [tex]\((x+9)^2\)[/tex] in the given equation. Therefore,
[tex]\[ x - (-9) = x + 9 \][/tex]
Hence, [tex]\(h = -9\)[/tex].
2. The term [tex]\((y-k)^2\)[/tex] in the standard form corresponds to [tex]\((y-6)^2\)[/tex] in the given equation. Therefore,
[tex]\[ y - 6 = y - 6 \][/tex]
Hence, [tex]\(k = 6\)[/tex].
Thus, the center [tex]\((h, k)\)[/tex] of the circle is:
[tex]\[ (-9, 6) \][/tex]
Therefore, the correct answer is:
[tex]\[ (-9, 6) \][/tex]
[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 of the circle.
Comparing the given equation [tex]\((x+9)^2 + (y-6)^2 = 10^2\)[/tex] with the standard form [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex], we can determine the values of [tex]\(h\)[/tex] and [tex]\(k\)[/tex].
1. The term [tex]\((x-h)^2\)[/tex] in the standard form corresponds to [tex]\((x+9)^2\)[/tex] in the given equation. Therefore,
[tex]\[ x - (-9) = x + 9 \][/tex]
Hence, [tex]\(h = -9\)[/tex].
2. The term [tex]\((y-k)^2\)[/tex] in the standard form corresponds to [tex]\((y-6)^2\)[/tex] in the given equation. Therefore,
[tex]\[ y - 6 = y - 6 \][/tex]
Hence, [tex]\(k = 6\)[/tex].
Thus, the center [tex]\((h, k)\)[/tex] of the circle is:
[tex]\[ (-9, 6) \][/tex]
Therefore, the correct answer is:
[tex]\[ (-9, 6) \][/tex]
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