I have:
$$x=5+8 \cos \theta$$ $$y=4+8 \sin \theta$$
With $ -\frac {3\pi}4 \le \theta \le 0$
If I wanted to write that implicitly, how would I do it? I get that it's a circle, and I can easily write the circle implicitly, but I'm not sure how to convert the domain from $ -\frac {3\pi}4 \le \theta \le 0$ to a suitable domain in x and y.
I can think of a way to do it that would involve using the $Min$ function, but is that allowed in an implicit representation?
On
Use that $\exists \theta : (v,w) = (\cos\theta, \sin\theta) \iff v^2 + w^2 = 1$.
You immediately get that the parametric curve is a part of the curve defined by the implicit equation $$ \left(\frac{x-5}8\right)^2 + \left(\frac{y-4}8\right)^2 = 1$$
To get only the part that you want, you must also make sure that $-\frac{3\pi}4\le \theta\le 0$, which is equivalent to $$ \cos \theta \ge -\frac 12, \sin\theta< 0 $$
and then you get the equation
\begin{cases} \left(\frac{x-5}8\right)^2 + \left(\frac{y-4}8\right)^2 = 1\\ \frac{x-5}8 \ge -\frac 12\\ \frac{y-4}8 \le 0 \end{cases}
Just compute the range of $x,y$, or using the geometric meaning of polar coordinate, draw a picture to see how $x,y$ changes on the circle when $\theta$ changes.
The result is $x\in[5-4\sqrt{2},13],y\in[-4,4]$, where $x,y$ satisfies $(x-5)^2+(y-4)^4=8^2$.