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Obtaining centre and radius of a circle from general equation of a circle

The equation ...

Question

The equation of a circle is $(x + 4)^{2} + y^{2} = 12$ . What is its center?

(- 4, 0)

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(4, 0)

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(0, 0)

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(0, - 4)

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Solution

The correct option is C $(- 4, 0)$
Compare the equation with the standard form with center at $(h, k)$ is: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Therefore, center of $(h, k) = (- 4, 0)$

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