# Sketch the curve described by $(x,y) = (3\cos t+2, 3\sin t -3)$

Sketch the curve described by these parametric equations. \begin{align} x &=3\cos t+2 \\ y &=3\sin t -3 \end{align} for $$0 \leq t < 2\pi$$.

I found the equation to be $$\left(\frac{y+3}{3}\right)^2+\left(\frac{x-2}{3}\right)^2 = 1$$ hence centre to be $$(2,-3)$$ and can sketch the graph on the cartesian plane but I don't know how to sketch it with this: $$0 \leq t < 2\pi$$.

Can someone explain and send an image of what it would look like? Thanks!

• Do you the cartesian equation of a circle? – hamam_Abdallah May 1 '20 at 2:57
• What? @hamam_Abdallah – Maths May 1 '20 at 2:58

Very straightforward, plug in values of $$t$$: just notice the offset:

Here are some points and the $$t$$ values that led to them:

Clear now??

• But what do I do for the 0 ≤ t < 2π I don't quite get how that will affect the graph and what changes I make. – Maths May 1 '20 at 3:00
• I have gotten that graph; that's easy but I don't get how to use the 0 ≤ t < 2π and how it's going to affect which parts of the graph are accepted and which aren't – Maths May 1 '20 at 3:01
• You USE $0 \leq t \leq 2 \pi$ to get the circle. Plug in $t=0$ to find one point on the plane; then plug in $t = 0.1$, then plug in $t=.2$... well, you get the idea. – David G. Stork May 1 '20 at 3:05

Your equation can be written as

$$(x-2)^2+(y+3)^2=9=3^2$$

your curve is then a circle with the point $$C(2,-3)$$ as center and Radius $$R=3$$.