Question

How do you write this equation `4/3x^2+4/3y^2=1?` in standard form and find the center and radius?

Answer 1

You can rewrite this as

`4/3x^2+4/3y^2=1=>4*(x^2+y^2)=3=>x^2+y^2=(sqrt3/2)^2`

Hence it is a circle with center at `(0,0)` and radius `r=sqrt3/2`