The standard form of the equation for a circle is
(x-a)^2 + (y-b)^2 = r^2
where
(a,b) = center of circle
r = radius.
So let's do it.
x^2 + y^2 + 8x + 22y + 37 = 0
Let's start with determining a. We know that (x-a)^2 will result in an x^2 term and an 8x term. Let's square (x-a) and we get x^2 - 2ax + a^2. The x^2 term is OK and we have
8x = -2ax
So
8x = -2ax
8 = -2a
-4 = a
So we now know that part of the equation will be (x - -4)^2 = (x + 4)^2. Let's expand that, giving x^2 + 8x + 16 and since we're adding 16 to the left, we need to add that to the right as well. So our equation becomes
(x + 4)^2 + y^2 + 22y + 37 = 16
Now let's work on the y terms. Using the same logic as above the equation to solve is
22y = -2by
22 = -2b
-11 = b
So (y - -11)^2 = (y+11)^2. And expanding gives us y^2 + 22y + 121 which means we need to add 121 to the right, so we have
(x + 4)^2 + (y + 11)^2 + 37 = 16 + 121
(x + 4)^2 + (y + 11)^2 + 37 = 137
(x + 4)^2 + (y + 11)^2 = 100
And a simple rewrite of the right hand side gives us
(x + 4)^2 + (y + 11)^2 = 10^2
So the final equation is
(x + 4)^2 + (y + 11)^2 = 10^2
Which indicates a circle with a radius of 10 centered at (-4,-11)
