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24 September, 14:28

The equation for the circle is:

x2+y2-12x+6y-19=0

What is the center of the circle?

(-6, 3)

(12, - 6)

(6, - 3)

(-12, 6)

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Answers (1)
  1. 24 September, 15:23
    0
    We have x^2 - 12x + 36 + y^2 + 6y + 9 = 64;

    Then, (x-6) ^2 + (y+3) ^2 = 6^2;

    The center of the circle is (6, - 3);
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