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Find the parametric equations for $x^2-4x+y^2-2y+5=2$, and graph. Hint: Complete some squares. I have completed squares and gotten $(x-2)^2+4=-(y+1)^2-2$ but I am confused with how to proceed. I know it will be a circle but how do I change this into parametric equations and graph points? They want the final form to be H(t)=(f(t), g(t))

Brynn
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hint: $$(x-2)^2 + (y+1)^2 = (\sqrt{2})^2\to x-2 = \sqrt{2}\cos \theta,y + 1 = \sqrt{2}\sin \theta$$.

If you have $\triangle^2 + \square^2 = A$, you can write $\left(\dfrac{\triangle}{\sqrt{A}}\right)^2+\left(\dfrac{\square}{\sqrt{A}}\right)^2 = 1$, then you can always set $\dfrac{\triangle}{\sqrt{A}} = \cos \theta$, and $\dfrac{\square}{\sqrt{A}} = \sin \theta$, for some angle $\theta \in [0,2\pi]$

DeepSea
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  • How did you get the sqrt(2)^2 and how did you know to turn that into the cos(theta)? – Brynn Mar 18 '15 at 02:12
  • you can rewrite the original equation as (x^2-4x +4) + (y^2-2y+1) = 2. or (x-2)^2 + (y-1)^2 = 2. remember cos^2(x) + sin^2(x) = 1. – user25406 Mar 18 '15 at 03:04
  • I'm very confused where the =2 came from...shouldn't it be -6? When I complete the square that's what I get. – Brynn Mar 18 '15 at 05:39
  • So how do I put this into the correct form and graph? I am struggling with the concepts behind this. – Brynn Mar 18 '15 at 05:50
  • So my f(t)=sqrt(2)cost and g(t)=sqrt(2)sint? – Brynn Mar 18 '15 at 06:01