Question

Convert the following rectangular equation to polar form.Assume a>0 3x^2+3y^2-4x+2y=0

Answer 1

The given equation is,

`3x^2+3y^2-4x+2y=0`

The polar form of the equation can be determined by using the substitution

`\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}`

using the substitution,

`\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}`

Thus, the above equation gives the required polar form of the circle.