SOLUTION: how to solve {{{x^2+4y^2+6x+16y+9=0}}} in standard form?

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Question 607124: how to solve x%5E2%2B4y%5E2%2B6x%2B16y%2B9=0 in standard form?
Found 2 solutions by lawdej, LisaJ:
Answer by lawdej(58) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+4y^2+6x+16y+9=0
use completing the square method
x^2+6x+4(y^2+4y)=-9
x^2+6x+9+4(y^2+4y+4)=-9+9+16
(x+3)^2+4(y+2)^2=16
(x+3)^2/16+4(y+8)^2/16=16/16
(x+3)^2/16+(y+8)^2/4=1 this is the standard form of x^2+4y^2+6x+16y+9=0(ellipse)

Answer by LisaJ(11) About Me  (Show Source):
You can put this solution on YOUR website!
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.
%28x%5E2+%2B+6x+%2B+9%29+%2B+%28y%5E2+%2B+16y+%2B+64%29+=+-9+%2B+9+%2B+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:
%28x%2B3%29%5E2+%2B+%28y%2B8%29%5E2+=+64+=+8%5E2
Comparing to standard form, I see this is a circle with radius 8, center at (-3, -8).