Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle.

Question

Find a parameterisation for $(x-8)^2 + y^2 = 16$ starting at the point $(4,0)$ and moving clockwise once around the circle.

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My work

We can describe the circle using polar coordinates:

$x(\theta) = \rho\cos(\theta)$, $y(\theta) = \rho\sin(\theta).$

$(x-8)^2 + y^2 = 16$ is a circle with radius of $4$.

$\therefore \rho = 4$

Let $t = \theta \ \forall \ t \in [0, 2\pi]$ be our parameter.

$\therefore (x(t), y(t)) = (4\cos(t), 4\sin(t))$

$(x(t), y(t)) = (4\cos(\pi-t), 4\sin(\pi-t))$ represents the circle of radius $4$ and starting at $\pi$ and moving clockwise.

However, this circle is still centred at $(0, 0)$. To centre the circle at $(8,0)$, we must have the parameterisation $(x(t), y(t)) = (8 + 4\cos(\pi - t), 4\sin(\pi-t)) \ \forall \ t \in [0, 2\pi]$.

We can test values some values to get an indication of any errors in our parameterisation.

$(x(0), y(0)) = (4, 0)$

$(x(\pi/2), y(\pi/2)) = (8, 4)$

$(x(\pi), y(\pi)) = (12, 0)$

$(x(3\pi/2), y(3\pi/2)) = (8, -4)$

$(x(2\pi), y(2\pi)) = (4, 0)$

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I would greatly appreciate it if people could please take the time to review my reasoning and solution for correctness.

Answer 1

Yes, the work is correct.

Below is a parametric plot of $\left\{8 + 4 \cos \left( \pi - t \right), 4 \sin \left(t\right)\right\}$ for $0\le t < 2\pi$.

To show that your parameterization starts at $(4,0)$ and proceeds in a clockwise manner, a sequence of values was created: $$ \begin{array}{cccc} k & t & x(t) & y(t) \\\hline 1 & 0 & 4 & 0 \\ 2 & \frac{\pi }{6} & 8-2 \sqrt{3} & 2 \\ 3 & \frac{\pi }{3} & 6 & 2 \sqrt{3} \\ 4 & \frac{\pi }{2} & 8 & 4 \\ 5 & \frac{2 \pi }{3} & 10 & 2 \sqrt{3} \\ 6 & \frac{5 \pi }{6} & 8+2 \sqrt{3} & 2 \\ 7 & \pi & 12 & 0 \\ 8 & \frac{7 \pi }{6} & 8+2 \sqrt{3} & -2 \\ 9 & \frac{4 \pi }{3} & 10 & -2 \sqrt{3} \\ 10 & \frac{3 \pi }{2} & 8 & -4 \\ 11 & \frac{5 \pi }{3} & 6 & -2 \sqrt{3} \\ 12 & \frac{11 \pi }{6} & 8-2 \sqrt{3} & -2 \\ 1 & 0 & 4 & 0 \\ 2 & \frac{\pi }{6} & 8-2 \sqrt{3} & 2 \\ 3 & \frac{\pi }{3} & 6 & 2 \sqrt{3} \\ 4 & \frac{\pi }{2} & 8 & 4 \\ 5 & \frac{2 \pi }{3} & 10 & 2 \sqrt{3} \\ 6 & \frac{5 \pi }{6} & 8+2 \sqrt{3} & 2 \\ 7 & \pi & 12 & 0 \\ 8 & \frac{7 \pi }{6} & 8+2 \sqrt{3} & -2 \\ 9 & \frac{4 \pi }{3} & 10 & -2 \sqrt{3} \\ 10 & \frac{3 \pi }{2} & 8 & -4 \\ 11 & \frac{5 \pi }{3} & 6 & -2 \sqrt{3} \\ 12 & \frac{11 \pi }{6} & 8-2 \sqrt{3} & -2 \\ \end{array} $$ The $k$ values are plotted against the circle.