$ABC$ is a straight line with $AB = BC = 3$ units. $B$ is the centre of the circle with radius of $2$ units. $P$ is a point on the circle. $\widehat{B_1} = \theta$, $\widehat{A} = x$, $QC \perp AC$ and $QC = y$.
Show that $$y = \frac{12\sin\theta}{3 + 2\cos\theta}$$.
How do I go about doing this? As you can see, I worked out as much as I can from the diagram and $$\sin\widehat{P_1} = \frac{3}{2} \times \sin x$$ I used the sine rule to work that out.
How do I get the required answer? Many thanks
Hint: You can use coordinate-geometry to get your answer. Put $B$ as the origin and $ABC$ as $x$-axis. Then $P\equiv(2\cos\theta,2\sin\theta)$
Can you take it from there?