Question 139738
The most useful form of the equation of a circle is:
  ( x - h )^2 + ( y - k )^2 = r^2
When the equation of a circle is written this way, the center of the circle can be easily read: (h, k). And the radius, r, can also be read easily.

So to solve your problem we need to use Algebra to change your original equation:
  x^2 + 12x + 36 + ( y - 4 )^2 = 49
to one that looks like:
  ( x - h )^2 + ( y - k )^2 = r^2

As I hope you can see, your original equation already has "y" in the proper form. All we need to do is change 
  x^2 + 12x + 36
to something like
  ( x - h )^2
and change whatever number is on the right to something like
  r^2

To change 
  x^2 + 12x + 36
to something like
  ( x - h )^2
we do something called "completing the square".

For
  x^2 + 12x + 36
completing the square happens to be very easy.
  x^2 + 12x + 36
is equal to 
  ( x + 6 )^2
(If this is not obvious, multiply ( x + 6 ) ( x + 6 )  [Most people use "FOIL" for this.] and see that
   x^2 + 12x + 36 = ( x + 6 )^2

Now ( x + 6 )^2 is not quite in the form ( x - h )^2. We need to rewrite it as a subtraction: ( x + 6 )^2 = ( x - (-6))^2

The only thing remaining is to rewrite the 49 on the right side as something squared. Since 49 = 7^2 this is pretty simple.

So our final version of the equation is
  ( x - (-6))^2 + ( y - 4 )^2 = 7^2

The center of the circle is (-6, 4) and the radius is 7.