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The centre of the circle x^2 + y^2 - 6x + 4y - 12 = 0 is
A(3, 2)
B(6, -4)
C(-3, 2)
D(3, -2)
Answer & Solution
Correct answer: D. (3, -2)
Completing squares gives (x-3)^2 + (y+2)^2 = 25, so the centre is (3, -2).
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