I solved it by first finding the Cartesian equation of $C$ and its domain & range.
$$
(x+2)^{2}+(y-1)^{2}=81,\;\;-2 \leq x \leq 7,\;\;-\frac{7}{2} \leq y \leq 10
$$
After sketching the curve we can see that the angle subtended by an arc $C$ at the centre is equal to $
\frac{\pi}{2}+\theta
$.
I then found the distance between the two marked points on the circle, and used the cosine law to find the angle $\theta$. Of course, $$
\begin{aligned}
\text {arc length C } &=\theta r \\
&=\left(\frac{\pi}{2}+\frac{\pi}{6}\right) \times 9 \\
&=6 \pi
\end{aligned}
$$
The answer booklet seems to suggest there is a much faster way of finding $\theta.$
If this is the case, could someone enlighten me please?
Best Answer
The parametric system defines a circle with radius $r=9.$ The parametrization is defined by the angle with vertex at the center of the circle, therefore the coordinates of the center do not influence the computation.
The interval $-\frac{\pi}{6}\leq t \leq \frac{\pi}{2}$ has a length $$l=\frac{\pi}{2}+\frac{\pi}{6}=\frac{2\pi}{3},$$ which corresponds to one third of a circle. Therefore, the length of $C$ is $$l(C)=r\times l =6\pi.$$