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