Question

what is the center and the radius of the following circle: x^2+y^2-8y=0

Answer 1

The equation for a circle is \(r^2=(x-a)^2+(y-b)^2\) where the center is the point \((a,b)\)
We want to get your form into this form, so we need to complete the square on the "y terms".

leaving the \(x^2\) term alone the entire time, we get

\(x^2+y^2-8y=0\implies x^2+(y-4)^2-16=0 \implies (x-0)^2+(y-4)^2=4^2 \)