circle eqauation - can ne1 double check my work

[Agent_J]

Find the center and radius of the circle
16x^2 + 16y^2 + 8x + 32y + 1 = 0

So first i simplified the equation by taking out the 16
so i got:

16 (x^2 + 1/2x + y^2 + 2y) = -1
16 (x^2 + 1/2x + 1/16) + 16 (y^2 + 2y + 1) = -1 + 1 + 1
16 (x + 1/4)^2 + 16 (y + 1)^2 = 1

Center = (-1/4, -1)
Radius = 1

Are my calculations correct? Do I need to take out the 16 in my equation?

[Muzza]

I believe you should write it in the form (x + 1/4)^2 + (y + 1)^2 = 1/16 = (1/4)^2. Then you can see that the radius is 1/4.

[Agent_J]

uh oh, then I must have done something wrong because the answer for the Radius should be just 1 :frown:

[Integral]

16 (x^2 + 1/2x + 1/16) + 16 (y^2 + 2y + 1) = -1 + 1 + 1

On the left hand side you added 16 on the right hand side you added 1.
If you add 16 to both sides it will work out.