Question

The equation for the circle is:

x2+y2−12x+6y−19=0
What is the center of the circle?
(-6, 3)
(12, -6)
(6, -3)

(-12, 6)

Answer 1

We have x^2 - 12x + 36 + y^2 + 6y + 9 = 64;
Then, (x-6)^2 + (y+3)^2 = 6^2;
The center of the circle is (6, -3);