Question

What is the standard form of the equation of the circle x^2 - 4x +y^2 +6y + 4 = 0 ?

Answer 1

Answer: (x-2)^2+(y+3)^2 = 9

Side notes
1) This circle has a center of (2,-3)
2) The radius of this circle is 3
3) The graph is shown in the attached image

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Work Shown:

x^2-4x+y^2+6y+4=0
x^2-4x+y^2+6y+4-4=0-4
x^2-4x+y^2+6y = -4
x^2-4x+4+y^2+6y = -4+4 ... see note 1 below
(x^2-4x+4)+y^2+6y = 0
(x-2)^2+y^2+6y = 0
(x-2)^2+y^2+6y+9 = 0+9 ... see note 2 below
(x-2)^2+(y^2+6y+9) = 9
(x-2)^2+(y+3)^2 = 9

note 1: I'm adding 4 to both sides to complete the square for the x terms. You do this by first taking half of the x (not x^2) coefficient which in this case is -4. So take half of -4 to get -2. Then square this result to get 4

note 2: Like with note 1, I'm completing the square. What's different this time is that this is for the y terms now. The y coefficient is 6. Half of this is 3. Square 3 to get 9. So this is why we add 9 to both sides.

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So the equation in standard form is (x-2)^2+(y+3)^2 = 9

Note how
(x-2)^2+(y+3)^2 = 9
is equivalent to 
(x-2)^2+(y-(-3))^2 = 3^2

So that second equation listed above is in the form (x-h)^2+(y-k)^2 = r^2
where
h = 2
k = -3
r = 3

making the center to be (h,k) = (2,-3) and the radius to be r = 3

The graph is attached.