The equation of a circle is given by x^2 + y^2 - 6x + 4y - 12 = 0. What are the coordinates of the center and the radius of the circle?
Correct: A
To find the center and radius, we need to rewrite the equation in standard form (x - h)^2 + (y - k)^2 = r^2 by completing the square.
Group the x-terms and y-terms: (x^2 - 6x) + (y^2 + 4y) = 12
Complete the square for x: (x^2 - 6x + (-6/2)^2) = (x^2 - 6x + 9)
Complete the square for y: (y^2 + 4y + (4/2)^2) = (y^2 + 4y + 4)
Add the constants added to both sides of the equation:
(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4
(x - 3)^2 + (y + 2)^2 = 25
From the standard form, the center (h, k) is (3, -2) and r^2 = 25, so the radius r = sqrt(25) = 5.