Center and radius of x^2 + y^2 – 4y = 0.

Answer 1

STD Form of Circle:
(x - h)^2 + (y - k)^2 = r^2, C(h , k).

x^2 + y^2 - 4y = o,
Complete the square:
x^2 + y^2 - 4y + (-4 / 2)^2 = (-4/2)^2,
x^2 + y^2 - 4y + 4 = 4,
Convert the perfect sq to a binomial:
(x - 0 )^2 + (y - 2)^2 = 4 = r^2, r=2.
C(0 , 2).