Answer
$(x-1)^{2}+(y+2)^{2}=1,\ \ 1\leq x\leq 2$
Work Step by Step
From the parametric equation for x, we see that $x\in[1,2].$
(sine is positive in quadrants I and II, that is, when $t\in[0,\pi]$)
From the parametric equation for $y$, we see that $y\in[-3,-1].$
(cosine varies from 1 to -1 in the first two quadrants)
Rewriting the parametric equations, isolating $\sin t$ and $\cos t,$
$\left\{\begin{array}{l}
x-1=\sin t\\
y+2=\cos t
\end{array}\right.$
Square both equations and add:
$ \fbox{$(x-1)^{2}+(y+2)^{2}=1,\ \ 1\leq x\leq 2$}$
- half of a circle of radius 1, centered at $(1,-2)$
To graph, create a table using several values for t, and then calculate $(x(t), y(t)),$ plotting the points as you go. Join with a smooth curve, noting the direction in which $t$ increases.
