Answer
(a) $x = 3~cos~t$
$y = 3~sin~t$
We can see the graph below.
(b) $x^2+y^2 = 9$
Work Step by Step
(a) $x = 3~cos~t$
$y = 3~sin~t$
When $t = 0$:
$x = 3~cos~0 = 3$
$y = 3~sin~0 = 0$
When $t = \frac{\pi}{6}$:
$x = 3~cos~\frac{\pi}{6} = \frac{3\sqrt{3}}{2}$
$y = 3~sin~\frac{\pi}{6} = \frac{3}{2}$
When $t = \frac{\pi}{4}$:
$x = 3~cos~\frac{\pi}{4} = \frac{3\sqrt{2}}{2}$
$y = 3~sin~\frac{\pi}{4} = \frac{3\sqrt{2}}{2}$
When $t = \frac{\pi}{3}$:
$x = 3~cos~\frac{\pi}{3} = \frac{3}{2}$
$y = 3~sin~\frac{\pi}{3} = \frac{3\sqrt{3}}{2}$
When $t = \frac{\pi}{2}$:
$x = 3~cos~\frac{\pi}{2} = 0$
$y = 3~sin~\frac{\pi}{2} = 3$
When $t = \frac{2\pi}{3}$:
$x = 3~cos~\frac{2\pi}{3} = -\frac{3}{2}$
$y = 3~sin~\frac{2\pi}{3} = \frac{3\sqrt{3}}{2}$
When $t = \pi$:
$x = 3~cos~\pi = -3$
$y = 3~sin~\pi = 0$
When $t = \frac{4\pi}{3}$:
$x = 3~cos~\frac{4\pi}{3} = -\frac{3}{2}$
$y = 3~sin~\frac{4\pi}{3} = -\frac{3\sqrt{3}}{2}$
When $t = \frac{3\pi}{2}$:
$x = 3~cos~\frac{3\pi}{2} = 0$
$y = 3~sin~\frac{3\pi}{2} = -3$
We can see the graph below.
(b) $x = 3~cos~t$
$y = 3~sin~t$
$x^2+y^2 = (3~cos~t)^2+(3~sin~t)^2$
$x^2+y^2 = 9~cos^2~t+9~sin^2~t$
$x^2+y^2 = 9~(cos^2~t+~sin^2~t)$
$x^2+y^2 = 9~(1)$
$x^2+y^2 = 9$
