Question 607124
To put the equation in standard form, the method is to 'complete the squares'. That means to collect all the x-terms in one place, and all the y-terms in another place. Then add a constant to each collection, so it makes a perfect square.

{{{(x^2 + 6x + 9) + (y^2 + 16y + 64) = -9 + 9 + 64}}}

I know I have to add 9 to the x-terms because that makes a perfect square. 6/2 = 3; 3^2 = 9.  Likewise for y-terms, 16/2 = 8; 8^2=64.  I also have to add all these constants to the other side of the equation, to keep it in balance.

Now I can factor each term:
{{{(x+3)^2 + (y+8)^2 = 64 = 8^2}}}
Comparing to standard form, I see this is a circle with radius 8, center at (-3, -8).