This is extremely similar to this question, but as there is no r next to the constant 1, when I multiply everything by r I'm going to end up with:
$r^2 = 4r + r\frac{\sin(\theta)}{2}$
And I don't see how to cancel out the 4r with the $x^2+y^2 = r$ identity.
How should I approach this?
And while I'm at it, how should I approach $r = \sin^2\theta?$ Again, multiplying by r would leave me with $r^2 = r\sin^2\theta$ and would still have a theta remaining once I pulled out $r\sin\theta$.
Assuming $r>0,$
$$2r^2=8r+r\sin\theta\implies2(x^2+y^2)-y=8r=8\sqrt{x^2+y^2}$$
squaring both sides we get $$(2(x^2+y^2)-y)^2=64(x^2+y^2)$$
Mind you the original relationship needs $2(x^2+y^2)-y=8r>0$