x² + 12x + y² + 12y + 36 = 0

That's the equation of a circle because it 
contains both an x² and y² with the same coefficient
when on the same side of the equation.

We must get it in standard form, which is

             (x - h)² + (y - k)² = r²

where the center is (h,k) and the radius is r. 

You must memorize that standard form and those facts
about h, k, and r.

Starting with 

        x² + 12x + y² + 12y + 36 = 0

Get the constant term 36 off the left side
Subtract 36 from both sides of the equation:

             x² + 12x + y² + 12y = -36 

Complete the square on the x's:

Take half of 12, the coefficient of x,
which is 6.  Then square 6 and get 36.
Now add 36 to both sides:

        x² + 12x + 36 + y² + 12y = -36 + 36 

Complete the square on the y's:

Take half of 12, the coefficient of y,
which is 6.  Then square 6 and get 36.
Now add 36 to both sides:

   x² + 12x + 36 + y² + 12y + 36 = -36 + 36 + 36

Factor the first three terms as (x + 6)(x + 6) and as (x + 6)²
and combine the terms on the right

        (x + 6)² + y² + 12y + 36 = 36

Factor the next three terms as (y + 6)(y + 6) and as (y + 6)²

        (x + 6)² + (y + 6)² = 36 

Now we compare with the standard form:

        (x - h)² + (y - k)² = r²

And see that -h = +6 or h = -6, -k = 6, so k = -6, and r² = 36
so r = 6.

So the equation is that of a circle with radius 6 and 
center (-6,-6)



Edwin