what would x^2+y^2-4x+2y=0 (Equation for a cirlce) be in
r^2=(x-h)^2(y-k)^2 form?
tnx
I will start you off...
x^2+y^2-4x+2y=0
x^2-4x+y^2 +2y =0
x^2-4x+ 4 + y^2 +2y+ 4 =+ 4+ 4
(x^2-4x+ 4) + (y^2 +2y+ 4) =8
To rewrite the equation x^2 + y^2 - 4x + 2y = 0 in the form r^2 = (x - h)^2 + (y - k)^2, we need to complete the square for both x and y.
Let's first complete the square for x:
x^2 - 4x + 4 = (x - 2)^2
Now, let's complete the square for y:
y^2 + 2y + 1 = (y + 1)^2
Substituting these completed square terms back into the equation, we get:
(x - 2)^2 + (y + 1)^2 = 8
Therefore, the equation x^2 + y^2 - 4x + 2y = 0 can be written in the form r^2 = (x - h)^2 + (y - k)^2 as:
r^2 = (x - 2)^2 + (y + 1)^2, where h = 2 and k = -1.