Question

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

Thank you! :)

Answer 1

In polar coordinates: x and y are equal to?

Answer 2

x= rcost
y=rsint

Answer 3

Still can't find the parametrization

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

Hmm could you tell me the way?