complete the square
x^2-4x+y^2+6y=0
x^2-4x+y^2+6y=0
x^2-4x + y^2+6y = 0
x^2-4x+4 + y^2+6y+9 = 4+9
(x-2)^2 + (y+3)^2 = 13
To complete the square for the equation x^2 - 4x + y^2 + 6y = 0, we will need to rearrange the equation and manipulate it in a way that allows us to create a perfect square trinomial for the x and y terms separately.
Step 1: Rearrange the equation by grouping the x-terms and the y-terms separately:
(x^2 - 4x) + (y^2 + 6y) = 0
Step 2: For the x-terms, we want to take half of the coefficient of x (-4) and square it. Then we add that result to both sides of the equation:
(x^2 - 4x + 4) + (y^2 + 6y) = 0 + 4
The equality remains true since we added 4 to both sides of the equation.
Step 3: Repeat the same process for the y-terms. We take half of the coefficient of y (6) and square it, then add the result to both sides of the equation:
(x^2 - 4x + 4) + (y^2 + 6y + 9) = 0 + 4 + 9
Again, the equality remains true since we added 9 to both sides of the equation.
Step 4: Factor the perfect square trinomials for both x and y terms and simplify:
(x - 2)^2 + (y + 3)^2 = 13
Now the equation is in the standard form of a circle, where the center of the circle is (2, -3) and the radius is the square root of 13.
So the completed square form of the given equation is: (x - 2)^2 + (y + 3)^2 = 13.