## University Calculus: Early Transcendentals (3rd Edition)

$x=2- \cos \theta$;$y=\sin \theta$; $0\leq \theta \leq 2 \pi$
The equation is a circle shifted from $(0,0)$ to $(2,0)$. Thus, we have: $y=\sin \theta$ $x=2- \cos \theta$ Hence, $x=2- \cos \theta$; $y=\sin \theta$; $0\leq \theta \leq 2 \pi$