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# translations in polar coordinates? watch

1. how would I translate something like r=2 so that the centre of the circle formed would be somewhere like (2,pi/4)?
2. It's probably simplest to convert it to Cartesians, apply the translation and then convert back to polars; you won't have much luck going straight from to the new equation (which is, I warn, a bit more complicated).
3. I've still got an r floating about that i can't get rid of
i tried getting the circle with centre (-1,-2) and radius 2 into polars and i get:
Code:
(x-1)^2 + (y-2)^2 = 2
x^2 - 2x +1 + y^2 - 4y + 4 = 2
r^2 -2x -4y + 5 = 2
r^2 - 2rcos(th) -4rsin(th) + 3 = 0
r^2 = 2rcosth + 4rsin(th) -3
r= 2cos(th) +4sin(th) -3/r
4. So, r^2 -(2cos(th)+4 sin(th)r + 3 = 0.

This is a quadratic in r, so you can solve it to write r as a function of th.
5. (Original post by Misiak)
I've still got an r floating about that i can't get rid of
i tried getting the circle with centre (-1,-2) and radius 2 into polars and i get:
Code:
(x-1)^2 + (y-2)^2 = 2
x^2 - 2x +1 + y^2 - 4y + 4 = 2
r^2 -2x -4y + 5 = 2
r^2 - 2rcos(th) -4rsin(th) + 3 = 0
r^2 = 2rcosth + 4rsin(th) -3
r= 2cos(th) +4sin(th) -3/r
If the argument of the centre is and the radius is 2 then the centre is rather than (-1,-2). You seem to have got the equation wrong in addition to this: remember that the equation of a circle with radius c and centre (a,b) is given by

You seem to have done , which definietly isn't right. It's funny (and frustrating) how quickly C1 knowledge disappears!

Once you've fixed this, do what DFranklin said.
6. I'm not sure whether this is the right time to comment that it seems pretty unlikely you'll get any kind of "pleasant" equation here. No-one would use polar coordinates in a scenario like this unless they had to.

(I am reminded of a bit in one of Feynman's books where they were designing something to calculate rocket trajectories during the war, and because they didn't discuss things well enough with the military people, didn't realise that the observer and the rocket launcher didn't have to be in the same place. Which meant introducing an (x, y) offset into the calculations. To paraphrase Feynmann "If we'd been using cartesian coordinates, it would have been fine. But everything was based in polar coordinates, and so this change was a big big mess".
7. yeah i'm realising this now (i also remember reading that book )
wikipedia has a general equation as

but my graphing software doesn't like it and just seems to give me spirals, gah

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Updated: September 29, 2010
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