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What is the standard form of the equation of the circle x2 -4x + y2 + 6y + 12 = 0?
4 months ago
Q:
What is the standard form of the equation of the circle x2 -4x + y2 + 6y + 12 = 0?
Accepted Solution
A:
We take the equation
x^2 -4x + y^2 + 6y + 12 = 0
and complete the square for x and y.
x^2 -4x + 4 - 4 + y^2 + 6y + 9 - 9 + 12= 0
(x-2)^2 - 4 + (y+3)^2 - 9 = -12
(x-2)^2+ (y+3)^2 = 1
Therefore the answer is (x-2)^2+ (y+3)^2 = 1