SOLUTION: Find the radius of a circle given by the following equation: x^2 + y^2 + 2x + 4y + 4 = 0 I know the answer (r=1), but I want to understand the steps to arrive at the solution

    x² + y² + 2x + 4y + 4 = 0

Swap the 2nd and 3rd terms on the left to
get the x-term next to the x² term and the 
y term next to the y-term, and get the 
constant off the left side by adding -4
to boith sides:

        x² + 2x + y² + 4y = -4

Skip a space after the 2x and after the 4y:

x² + 2x + _ + y² + 4y + _ = -4

1. Multiply the coefficient of x, which is 2, by , getting 1
2. Square 1, getting 1², which is 1
3. Add that 1 in the first blank, and also add +1 to the right side

x² + 2x + 1 + y² + 4y + _ = -4 + 1

1. Multiply the coefficient of y, which is 4, by , getting 2
2. Square 2, getting 2², which is 4
3. Add that 4 in the first blank, and also add +4 to the right side
 
x² + 2x + 1 + y² + 4y + 4 = -4 + 1 + 4

Factor the first three terms on the left as (x+1)(x+1) or (x+1)²

   (x + 1)² + y² + 4y + 4 = -4 + 1 + 4

Factor the last three terms on the left as (y+2)(x+2) or (y+2)²

      (x + 1)² + (y + 2)² = -4 + 1 + 4

Combine the terms on the right

      (x + 1)² + (y + 2)² = 1

Compare to standard equation for a circle which you should
memorize as: 

      (x - h)² + (y - k)² = r²

-h = 1, -k = 2, r² = 1 so
 h = -1, k = -2, r = 1

So the center of the circle is (h,k) = (-1,-2) and the radius is r=1
Here's the graph of that circle:



Edwin