Question

x^2+y^+3x+4y+4=0 How do you get the standard form of the circle?

Answer 1

you must complete the square for both x and y

x^2 + 3x = (x + 3/2)^2 - 9/4 [if you expand it you'll see it's true]
y^2 + 4y = (y + 2)^2 - 4

whole thing is now

(x + 3/2)^2 - 9/4 + (y + 2)^2 - 4 + 4 = 0

(x + 3/2)^2 + (y + 2)^2 = 9/4



if you're unclear about completing the square:

(x + a)^2 = x^2 + 2a + a^2

so x^2 + 2a = (x + a)^2 - a^2