Question

4x^2+4y^2+36y+5=0 how do I put it in standard form

Answer 1

4x^2+4y^2+36y+5=0
... x^2 and y^2 with equal coefficient -> circle with standard form (x-h)^2 + (y-k)^2 = r^2
... group x, y and constant
(4x^2) + (4y^2 + 36y) = -5
... eliminate x,y coefficient (div by 4)
x^2 + (y^2 + 9y) = -5/4
... add necessary term to form
x^2 + 2ax + a^2 = (x + a)^2

x^2 + [y^2 + 2(9/2)y] = -5/4
... need (9/2)^2 for y-term
x^2 + [y^2 + 2(9/2)y + (9/2)^2] = -5/4 + (9/2)^2
x^2 + (y + 9/2)^2 = 19 = (sqrt(19))^2