By drawing a diagram, we can more clearly identify the curve $D$.

We first observe that the points O,Q,P are collinear, (Q is the point of intersection of the line OP with the circle $C$). We next note that the points A,Q and P make an isosceles triangle (which is precisely the property that defines the curve C!) Expressions for the lengths $AP$ and $QP$ can now be determined $$|AP| = \sqrt{(x-3)^2 + y^2}$$ $$|QP| = \sqrt{(x-\cos\theta)^2 + (y -\sin \theta)^2}$$ The point $P$ is such that it satisfies $$ \begin{array}{ccc} |AP| &=& |QP| \\ \sqrt{(x-3)^2 + y^2}& = & \sqrt{(x-\cos\theta)^2 + (y -\sin \theta)^2}\\ -6x + 9 &=& -2\cos x - 2\sin y + 1 \\ 2\cos\theta x + 2\sin \theta y &=& 6x - 8 \\ \end{array} $$ Since the points O,Q,P are collinear, then y we can see that $P$ must lie on the line $y=(\tan \theta) x$. From this, we can derive expressions for $\sin \theta$ and $\cos \theta$ in terms of $x,y$: $$\sin \theta = \frac{y}{\sqrt{x^2 + y^2}}, \cos\theta = \frac{x}{\sqrt{x^2+y^2}}$$ $$ \begin{array}{ccc} 2\cos\theta x + 2\sin \theta y &=& 6x - 8 \\ \frac{2x^2}{\sqrt{x^2+y^2}} + \frac{2y^2}{\sqrt{x^2+y^2}} &=& 6x-8 \\ \left(\frac{(x^2+y^2)}{3x-4}\right)^2 &=& x^2 + y^2 \end{array} $$