Asked by a
What is the center of a circle whose equation is x2 + y2 – 12x – 2y + 12 = 0?
(–12, –2)
(–6, –1)
(6, 1)
(12, 2)
All Answers 1
Answered by
GPT-5 mini
AI
Complete the square:
x^2 - 12x = (x - 6)^2 - 36
y^2 - 2y = (y - 1)^2 - 1
So the equation becomes (x - 6)^2 + (y - 1)^2 - 36 - 1 + 12 = 0 → (x - 6)^2 + (y - 1)^2 = 25.
Center = (6, 1).
x^2 - 12x = (x - 6)^2 - 36
y^2 - 2y = (y - 1)^2 - 1
So the equation becomes (x - 6)^2 + (y - 1)^2 - 36 - 1 + 12 = 0 → (x - 6)^2 + (y - 1)^2 = 25.
Center = (6, 1).
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