#### Answer

(a) $x = 2+sin~t$
$y = 1+cos~t$
We can see the graph below.
(b) $(x-2)^2+(y-1)^2 = 1$

#### Work Step by Step

(a) $x = 2+sin~t$
$y = 1+cos~t$
When $t = 0$:
$x = 2+sin~0 = 2$
$y = 1+cos~0 = 2$
When $t = \frac{\pi}{6}$:
$x = 2+sin~\frac{\pi}{6} = 2.5$
$y = 1+cos~\frac{\pi}{6} = 1+\frac{\sqrt{3}}{2}$
When $t = \frac{\pi}{4}$:
$x = 2+sin~\frac{\pi}{4} = 2+\frac{\sqrt{2}}{2}$
$y = 1+cos~\frac{\pi}{4} = 1+\frac{\sqrt{2}}{2}$
When $t = \frac{\pi}{3}$:
$x = 2+sin~\frac{\pi}{3} = 2+\frac{\sqrt{3}}{2}$
$y = 1+cos~\frac{\pi}{3} = 1.5$
When $t = \frac{\pi}{2}$:
$x = 2+sin~\frac{\pi}{2} = 3$
$y = 1+cos~\frac{\pi}{2} = 1$
When $t = \frac{2\pi}{3}$:
$x = 2+sin~\frac{2\pi}{3} = 2+\frac{\sqrt{3}}{2}$
$y = 1+cos~\frac{2\pi}{3} = 0.5$
When $t = \pi$:
$x = 2+sin~\pi = 2$
$y = 1+cos~\pi = 0$
When $t = \frac{4\pi}{3}$:
$x = 2+sin~\frac{4\pi}{3} = 2-\frac{\sqrt{3}}{2}$
$y = 1+cos~\frac{4\pi}{3} = 0.5$
When $t = \frac{3\pi}{2}$:
$x = 2+sin~\frac{3\pi}{2} = 1$
$y = 1+cos~\frac{3\pi}{2} = 1$
We can see the graph below.
(b) $x = 2+sin~t$
$x-2 = sin~t$
$y = 1+cos~t$
$y-1 = cos~t$
$(x-2)^2+(y-1)^2 =sin^2~t+cos^2~t$
$(x-2)^2+(y-1)^2 = 1$