What is the center and radius of the circle? *
x^2+ y^2+ 4y + 4 = 36

(0, - 2 ), r = 6

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

Rearrange the given equation collecting the terms in y together, that is

x² + y² + 4y + 4 = 36 ( subtract 4 from both sides )

x² + y² + 4y = 32

Use the method of completing the square on the y- terms

add ( half the coefficient of the y- term)² to both sides

x² + y² + 2(2)y + 4 = 32 + 4

x² + (y + 2)² = 36 ← in standard form

with (h, k ) = (0, - 2 ) and r² = 36, thus

centre = (0, - 2 ) and r = = 6

is only a line? i think so