[tex]r=\dfrac{12}{4+8\sin\theta}=\dfrac3{1+2\sin\theta}[/tex]
Let [tex]y=2\sin\theta[/tex], and recall that in polar coordinates, [tex]r=\sqrt{x^2+y^2}[/tex]. This means you have
[tex]\sqrt{x^2+y^2}=\dfrac3{1+y}[/tex]
You can stop there, or try to find something that looks somewhat nicer.
[tex]x^2+y^2=\dfrac9{(1+y)^2}[/tex]
