You'll need to group all the x terms together in one place and allt he y terms together in another place.
Standard equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where r is the radius and the point (h,k) is the center of the circle.
Rearranging x^2+y^2+6x+14y+37=0 produces
x^2 + 6x + y^2 + 14y = -37
We must now "complete the square" of x^2 + 6x.
Take half of the coefficient of x (which is 6) and square that. Add this square to x^2 + 6x, and then subtract this square:
x^2 + 6x + (3)^2 - (3)^2
Now do the same thing to the given y terms.
x^2 + 6x + 9 - 9 + y^2 + 14y + (7)^2 - (7)^2 = -37
This becomes This becomes
(x+3)^2 -9 (y+7)^2 - 49 = -37
Add 9 and add 49 to both sides. You end up with
(x+3)^2 + (y+7)^2 = 49 + 9 -37 = r^2
Find r.
This is the equation of the circle in standard form. Its center is at (-3,-7).
