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


Sagot :

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

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

Here, \((h, k)\) represents the coordinates of the center of the circle, and \(r\) is the radius.

Given the equation in the problem:

[tex]\[ (x + 9)^2 + (y - 6)^2 = 10^2 \][/tex]

Let's rewrite \((x + 9)^2\) in a form that matches the standard equation format. Notice that:

[tex]\[ (x + 9) = (x - (-9)) \][/tex]

So, \((x + 9)^2\) can be written as:

[tex]\[ (x - (-9))^2 \][/tex]

Similarly, the term \((y - 6)^2\) already matches the standard form.

Now, comparing:

[tex]\[ (x - (-9))^2 + (y - 6)^2 = 10^2 \][/tex]

we see that \((h, k)\) corresponds to \((-9, 6)\).

Therefore, the center of the circle is:

[tex]\[ \boxed{(-9, 6)} \][/tex]