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
