Draw the planar curve:
$$\begin{align} r(t) &= \langle 2\cos(t) + 1 ,\, - 2\sin(t) + 3 \rangle \quad\text{for }0 < t < \pi\\ r’ (t) &= \langle - 2\sin(t) ,\, - 2\cos(t) \rangle\\ r’ (t) &= \langle - 0.1 ,\, - 2 \rangle \end{align}$$
The curve is an ellipse but is it a circle?
I’m guessing it’s easier to plug in t values to determine $x$ and $y$ values
Since $(x-1)^2+(y-3)^2=2^2$, it's a centre-$(1,\,3)$ radius-$2$ circle - well, a semicircular arc, if we're restricted to $t\in(0,\,\pi)$ rather than $[0,\,2\pi)$.