Question

Convert (x-a)^2+y^2=k into polar co-ordinates

Answer 1

From 0,0

Answer 2

turn it into this form

\(r^2=2ar \cos \theta+k-a^2\)

? or .... actually i dont know what do u mean by from 0,0

Answer 3

Here is (x-a)^2+y^2=k
I'd like the 'centre of polar rotation' (i.e. the dot in the 1st picture) to be at the cartesian co-ordinates of 0,0

Answer 4

How do you get it into the form that you stated (sorry, I know next to nothing about polar co-ordinate making)

Answer 5

Converting between polar and Cartesian coordinates

\(x^2+y^2=r^2\)
\(x=r \cos \theta\)
\(y=r\sin \theta\)

Answer 6

the way you have it\[(x-a)^2+y^2=k\]the radius is\[r=\sqrt k\]so\[x=\sqrt k\cos\theta\]\[y=\sqrt k\sin\theta\]plugging this in gives\[(\sqrt k\cos\theta-a)^2+(\sqrt k\sin\theta)^2=k\]which you can simplify a bit

Answer 7

I mean\[(\sqrt k\cos\theta-a)^2+(\sqrt k\sin\theta)^2=r^2\]

Answer 8

I messed this up explanation before, let me try again
conversion to polar coordinates is\[x=r\cos\theta\]\[y=r\sin\theta\]and a circle is given by the formula\[x^2+y^2=r^2(\cos^2\theta+\sin^2\theta)=r^2=Constant\]now to shift this circle around the have a center \((h,k)\) all we need to do is make the appropriate adjustment for our changes for x and y:\[x=h+r\cos\theta\]\[y=k+r\sin\theta\]so the circle at center \((h,k)\) in polar is\[(h+r\cos\theta)^2+(k+r\sin\theta)^2=Constant\]so the constant is the square of the actual radius of the circle, and \(r\) is the radius from the origin that traces out the circle as \(\theta\) changes
I hope that made sense

Answer 9

How would you put that into the form r=f(θ)?

Answer 10

solve for r...

Answer 11

Rearrange it?

Answer 12

\[h^2+2rh\cos\theta+r^2+k^2+2rk\sin\theta=C\]I never claimed it to be easy...
and with r^2 it may not technically be a function upon solving for r

Answer 13

this sucker is quadratic in r and you could use the quadratic formula if you wished

Answer 14

I think the message here is that it kind of defeats the purpose of using polar coordinates to simplify things unless they have a polar symmetry about the origin of some kind
even though the graph has a rotational symmetry, the benefit is broken by shifting it and breaking the symmetry of the whole graph
resulting in something fairly ugly

Answer 15

no on the other hand if you leave it parametric and keep the coordinates separated it is fine to use\[x=h+r\cos\theta\]\[y=k+r\sin\theta\]\[\vec f(\theta)=\langle h+r\cos\theta,k+r\sin\theta\rangle\]

Answer 16

now on the....*

Answer 17

that is a parametric vector function that traces out the circle

Answer 18

My problem is that I'm trying to decompose r(θ) into its x and y components to work out the force on an object- damn, you had actually answered that earlier, but I was overcomplicating- Thanks and apologies for dragging this out!

Answer 19

no problem, a little context helps sometimes :)

Answer 20

Especially as I am useless at polar equations...

Answer 21

I am pretty lazy about them too
I consult things like this from time to time
http://tutorial.math.lamar.edu/Classes/CalcII/PolarCoordinates.aspx