- #1

- 7

- 0

(x-h)^2+(y-k)^2= h^2+k^2

It is a circle with k and h greater than 0.

- Thread starter Tarhead
- Start date

- #1

- 7

- 0

(x-h)^2+(y-k)^2= h^2+k^2

It is a circle with k and h greater than 0.

- #2

- 49

- 0

http://mathworld.wolfram.com/PolarCoordinates.html

I think the transforms would be

x--> rcos theta

y--> r sin theta

h --> R cos theta'

k --> R sin theta'

4 prameters to describe the points on a shifted circle (shifted orgin because of the k and h terms) in either cartesian or polar coordinates

Not sure, but I think.

- #3

James R

Science Advisor

Homework Helper

Gold Member

- 600

- 15

[tex]x = r\cos \theta, \qquad y = r\sin \theta[/tex]

That's all you need.

- #4

- 49

- 0

You would need four parameters to specify a shifted circle in either coordinate system. (The k and h parameters will propagate through your transformation.) You could transform this shift into polar coordinates as well (and you would have to if this was a complicated mechanics problem) but you don't even need to bother with this because it is given as a constant.

Hope I didn't mess you up. Sorry again.

- Replies
- 1

- Views
- 587

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 12

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 180

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 11

- Views
- 5K

- Last Post

- Replies
- 7

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 776

- Last Post

- Replies
- 1

- Views
- 791

- Last Post

- Replies
- 0

- Views
- 2K