Can you help me put these in standard form?

Question

The circle whose equation is x^2 + 2x + y^2 + 6y + 6 = 0
has center (1,3) and radius 2
has center (-1, -3) and radius 2
has center (-1, -3) and radius 4
has center (1, 3) and radius 4


The circle whose equation is x^2 - 6x + y^2 + 2y = -4
has center (-3,1) and radius sqrt(6)
has center (3,-1) and radius sqrt(6)
has center (3,-1) and radius 6
has center (-1,3) and radius sqrt(6)


COULD YOU ALSO SHARE HOW YOU GOT YOUR ANSWER! I am having a lot of difficulty with this

Answer 1

x^2 + 2x + y^2 + 6y = -6
x^2 + 2x + 1 + y^2 + 6y + 9 = -6 + 10
(x+1)^2 + (y+3)^2 = 4
center = (-1, -3) radius = 2

x^2 - 6x + y^2 + 2y = -4
x^2 - 6x + 9 + y^2 + 2y + 1 = -4 + 10
(x-3)^2 + (y+1)^2 = 6
center =(3,-1) radius = sqrt(6)

Answer 2

The standard form of circle is (x - a)^2 + (y - b)^2 = r^2

where center is (a,b) and r is radius

1)
x^2 + 2x + y^2 + 6y + 6 = 0

complete the squares by adding required values at both sides

x^2 + 2x +1 + y^2 + 6y + 9 + 6 = 1+9

(x + 1)^2 + (y+3)^2 + 6 = 10

(x + 1)^2 + (y+3)^2 = 4

(x + 1)^2 + (y+3)^2 = 2^2

the center is (-1, -3) and rsdius is 2

2)

x^2 - 6x + y^2 + 2y = -4

x^2 - 6x + 9 + y^2 +2y + 1 = -4 + 9 + 1

(x - 3)^2 + (y + 1)^2 = 6

so center is (3, -1) and r = sqrt(6)