The given equation is,
[tex]3x^2+3y^2-4x+2y=0[/tex]
The polar form of the equation can be determined by using the substitution
[tex]\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}[/tex]
using the substitution,
[tex]\begin{gathered} 3(x^2+y^2)-4x+2y=0 \\ 3(r^2\cos ^2\theta+r^2\sin ^2\theta)-4r\cos \theta+2r\sin \theta=0 \\ 3r^2-4rcos\theta+2r\sin \theta=0 \\ r(3r-4\cos \theta+2\sin \theta)=0 \\ r=0\text{ and }(3r-4\cos \theta+2\sin \theta)=0 \\ (3r-4\cos \theta+2\sin \theta)=0 \end{gathered}[/tex]
Thus, the above equation gives the required polar form of the circle.
