Write the equation in standard form x^2+y^2+6x+14y+37=0

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).