Could someone explaing how I got this answer

Transform the polar equation to an equation in rectangular coordinates. Then identify and graph the equation.

r = -6 cos θ

So here's my work..
(r) r = -6 cos θ (r)

r^2 = -6 r cos θ

x^2 + y^2 = -6 x
(+6x).........(+6x)

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

...part I don't understand was where this 9 came from..

(x^2 + 6x + 9) y^2 = 0

I eventually come up with the answer (x + 3)^2 + y^2 = 9

To complete the square you need to add 9 to x^2 + 6x
since x^2 + 6x + 9 = (x + 3)^2

BUT, you don't just add 9 to one side of equation.
To keep equality, you add 9 to BOTH sides of equation.

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

Adding 9 to both sides will give perfect square on left side and will NOT change the equation.

x^2 + 6x + 9 + y^2 = 0 + 9
(x + 3)^2 + y^2 = 9

Equation of circle centred at (−3, 0) with radius 3

—————————————————————

x^2 + (y^2 + 14y) = 0

To complete the square add (14/2)^2 = 49 to BOTH sides:

x^2 + (y^2 + 14y + 49) = 49

Now factor:

x^2 + (y + 7)^2 = 49

Equation of circle centred at (0, −7) with radius 7

The "9" is how you complete the square for x
(x^2 + 6x) + y^2 = 0
==> x^2 + 6x + 9 + y^2 = 9
==> (x + 3)^2 + y^2 = 9
For the other... (assuming you are right to that point)
complete the square for the y part
y^2 + 14 y + (14/2)^2 - (14/2)^2
==> y^2 + 14y + 49 - 49
==> (y + 7)^2 - 49
so
x^2 + (y + 7)^2 = 49