Question

The equation of a circle is x^2+y^2+4x+8y=12.
What is the center of the circle?

Answer 1

general eq. of the circle is
x^2 + y^2 + 2gx + 2fy + c = 0
and center of the circle is ( -g , -f )
so, center = ( -2, -4)

Answer 2

You will need to convert the equation to the centre, radius form.
x^2+y^2+4x+8y=12

You must gather the x's and the y's separately and complete the square for each

x^2+y^2+4x+8y=12
x^2+4x + y^2+8y=12 (move x's together, y's together)

Now we need to complete the square for both x and y.
a^2 + 2ab + b^2 = (a + b)^2
Remember to do this you find half of the coefficient of x or y, then square that and add it at the end.

x^2+4x+4 + y^2 + 8y + 16 = 12 + 4 + 16 (remember to balance both sides)
(x+2)^2 + (y + 4)^2 = 32 (rewrite as perfect square)

So the centre is (-2,-4) and the radius is sqrt(32) = 4 sqrt(2)