Question

How do you find the center and radius of the circle given `x^2+y^2-6y-16=0`?

Answer 1

The process is complete the squares. See below

Explanation:

Remember that `(a+b)^2=a^2+2ab+b^2` and
`(a-b)^2=x^2-2ab+b^2`

Then
`x^2+y^2-6x-16=(x-3)^2+y^2-16-9=(x-3)^2+y^2-25=0`

Then we have finally `(x-3)^2+y^2=25=5^2`

This is the equation of a circle centered in `(3,0)` and radius 5