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