Parametrize (x-3)^2 + y^2 = 1

Thank you! :)

Question

Parametrize (x-3)^2 + y^2 = 1

Thank you! :)

chrisasl · Answer

In polar coordinates: x and y are equal to?

anonymous · Answer

x= rcost
y=rsint

anonymous · Answer

Still can't find the parametrization

chrisasl · Answer

Sorry, can't think of the solution now :-/ If I think of something, I'll reply.
Good luck with this!

anonymous · Answer

Thank you! :) I'll ask my professor tomorrow, I'll post the solution once I have it, but I'll try to keep figuring it out

chrisasl · Answer

We could do that:$$x^2 + y^2 - 6*x + 9 = 1 $$
Or: $$x^2 + y^2 - 6*x = -8$$ after that we plug in the formulas for x and y, so:
$$r^2 - 6*r*\cos(t) = -8$$

But I'm not sure that this is correct.

Do as you said and reply with the solution after you ask your professor.

anonymous · Answer

ok so, I think I found it out :)
$$r(t) = 3+\cos(t), \sin(t)$$

chrisasl · Answer

Hmm could you tell me the way?