What is the center of the circle given by x^2 + y^2 - 4x + 6y - 3 = 0?
Show the answer and explanation
Correct answer
A. (2, -3)
Principle or equation
Complete the square to rewrite the circle equation in the form (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center.
Why this answer is correct
Group x and y terms: (x^2 - 4x) + (y^2 + 6y) = 3. Complete squares: (x - 2)^2 - 4 + (y + 3)^2 - 9 = 3, so (x - 2)^2 + (y + 3)^2 = 16. Center is (2, -3).
Example
For x^2 + y^2 - 2x + 4y - 4 = 0, completing squares gives (x - 1)^2 + (y + 2)^2 = 9, center (1, -2).
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