precalculus
posted by muffy .
Determine the center and the radius of the following circle:
x^2+y^2+8x2y=8
so I figured I'd use completing the square after putting it in standard form:
(x^2+8x)+(y^22y)=8
I'm not sure where to go from here.

but you didn't "complete" the square
(x^2+8x)+(y^22y)=8
(x^2+8x + 16)+(y^22y + 1)= 8 + 16 + 1
(x+4)^2 + (y1)^2 = 25
Now we have it in standard form.
Can you take it from there ? 
Ok. Do I take the square root of the whole thing now?
getting x+4y1+sqrt25
xy=3+sqrt25
I've never done this before, I'm not sure if this is even right and if it is, I definitely don't know where to go from here 
no,
the standard equation of a circle with centre (h,k) and radius r is
(xh)^2 + (yk)^2 = r^2
so we ended up with
(x+4)^2 + (y1)^2 = 25 or
(x+4)^2 + (y1)^2 = 5^2
then the centre is (4,1) and radius is 5
Here is an interesting page for your topic
http://www.analyzemath.com/CircleEq/Tutorials.html 
Thanks so much for your explanation. I was able to figure it out and now got your answer.
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