Could anyone help me with this

$x = 1 + \cos t$, $y = −2 + \sin t$, $π ≤ t ≤ 2π$;

$x = t$, $y = −2 −\sqrt{2t − t^2}$, $0 ≤ t ≤ 2$

For the following parametric equations, how do I determine whether they both represent the same curve? And how to represent the curve in a single equation? What I did is to sub in the value of x and y from both equations to find the value of t, but i don't know how to proceed. Or should i draw the curve out and check whether it is the same curve?

## Best Answer

$x = 1 + \cos t, \ \ \ y = −2 + \sin t, \ \ \ \pi \leq t \leq 2 \pi$:

$$(x-1)^2+(y+2)^2=(1 + \cos t-1)^2+(−2 + \sin t+2)^2=\cos^2 t+\sin^2 t=1 \\ \Rightarrow (x-1)^2+(y+2)^2=1$$

$x = t, \ \ \ y = −2 −\sqrt{2t − t^2}, \ \ \ 0 \leq t \leq 2$:

$$(x-1)^2+(y+2)^2=(t-1)^2+(−2 −\sqrt{2t − t^2}+2)^2=t^2-2t+1+2t-t^2=1 \\ \Rightarrow (x-1)^2+(y+2)^2=1$$

Therefore, the parametric equations represent the same curve.