Question

what is the standard form of the equation of the circle 3x^2+3y^2-12y-0?

Answer 1

i assume that is \(=0\) not \(-0\)

Answer 2

yes it is

Answer 3

divide by 3 first and start with
\[x^2+y^2-4y=0\] then complete the square for the \(y\) terms
do you know how to do that?

Answer 4

No I do not

Answer 5

what is half of \(4\) ?

Answer 6

2

Answer 7

ok then we start with
\[x^2+(y-2)^2=?\] now what is \(2^2\) ?

Answer 8

I thought it was a -4?

Answer 9

complete the square for y... (-4/2)^2

Answer 10

minus does not matter when you square it

Answer 11

half of \(-4\) is \(-2\) and then \(2^2=4\) so you get
\[x^2+(y-2)^2=4\]