Question 24555
Find the centre of the circle: {{{x^2 + 4x + y^2 + 2y - 9 = 0}}}
First get your equation into the standard form for a circle.  You can do this by "completing the square" in the x-terms and in the y-terms.
{{{(x^2 + 4x) + (y^2 + 2y - 9) = 0}}} Complete the square in the x- and y-terms by adding the square of half the x-, y-coefficients to both sides of the equation.
{{{(x^2 + 4x + 4) + (y^2 + 2y + 1)- 9 = 1+4}}} Factor and Simplify.
{{{(x + 2)^2 + (y + 1)^2 - 9 = 5}}} Add 9 to both sides.
{{{(x + 2)^2 + (y + 1)^2 = 14}}} Now compare this with the standard form for a circle with centre at (h, k).
{{{(x - h)^2 + (y - k)^2 = r^2}}}
You can see that the centre of the circle is at (-2, -1)