Vorsah
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I know that x=rcos(theta) and y=rsin(theta)

How would I change this to for different coordinates, for example say if I have a circle with centre (0,2) how would I change this to polar coordinates and also if I have a circle with centre (2,0) how would I change this?

Thanks
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Jai Sandhu
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(Original post by Vorsah)
I know that x=rcos(theta) and y=rsin(theta)

How would I change this to for different coordinates, for example say if I have a circle with centre (0,2) how would I change this to polar coordinates and also if I have a circle with centre (2,0) how would I change this?

Thanks
Construct a cartesian equations in terms of x and y for that circle, then sub in the 2 equations you initially quoted if you want cartesian -> polar.
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Vorsah
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(Original post by Jai Sandhu)
Construct a cartesian equations in terms of x and y for that circle, then sub in the 2 equations you initially quoted if you want cartesian -> polar.
What I mean is say if you have the point (0,1) then in polar form this would be x=rcostheta) and y=1+rsin(theta). So what would it was point (1,0)?
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Jai Sandhu
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(Original post by Vorsah)
What I mean is say if you have the point (0,1) then in polar form this would be x=rcostheta) and y=1+rsin(theta). So what would it was point (1,0)?
The second sentence makes no sense so I am just going to ignore that. The point 0,1 would be the same as:

x=rcostheta

and

y-1=rsin(theta) which is the same as y=1+rsin(theta)
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Vorsah
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What would the point (1,0) be?

Y=rsin(theta)
X=1+cos(theta) ?
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username1560589
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(Original post by Vorsah)
What would the point (1,0) be?

Y=rsin(theta)
X=1+cos(theta) ?
The point (1, 0) in polar is (1/\sqrt{2}, \pi/4).

To find the equation of a circle in polar form, construct a cartesian equation in the form (x-a)^2 + (y-b)^2 = c^2 and then substitute the equations at the top of the thread.
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Vorsah
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(Original post by Jai Sandhu)
The second sentence makes no sense so I am just going to ignore that. The point 0,1 would be the same as:

x=rcostheta

and

y-1=rsin(theta) which is the same as y=1+rsin(theta)
(Original post by morgan8002)
The point (1, 0) in polar is (1/\sqrt{2}, \pi/4).

To find the equation of a circle in polar form, construct a cartesian equation in the form (x-a)^2 + (y-b)^2 = c^2 and then substitute the equations at the top of the thread.

Im doing double integration and I've been given the domain: x^2 + (y-1)^2=1, so its centre 0,1 which in polar coordinates is x=rcos(theta) and y=1+rsin(theta).

So I want to know if the domain is: (x-1)^2 +y^2=1 with centre (1,0) what would this be in polar coordinates?
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username1560589
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(Original post by Vorsah)
Im doing double integration and I've been given the domain: x^2 + (y-1)^2=1, so its centre 0,1 which in polar coordinates is x=rcos(theta) and y=1+rsin(theta).
These are parametric, not polar equations.
So I want to know if the domain is: (x-1)^2 +y^2=1 with centre (1,0) what would this be in polar coordinates?
Are you asking what the domain would be?
If so, you can find the minimum and maximum values of \theta by sketching the curve and working the minimum and maximum angles geometrically.
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Vorsah
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(Original post by morgan8002)
These are parametric, not polar equations.
Are you asking what the domain would be?
If so, you can find the minimum and maximum values of \theta by sketching the curve and working the minimum and maximum angles geometrically.
I think I was getting confused between polar and parametric equations. What would the point (1,0) be in parametric equations?
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Jai Sandhu
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(Original post by morgan8002)
These are parametric, not polar equations.
Are you asking what the domain would be?
If so, you can find the minimum and maximum values of \theta by sketching the curve and working the minimum and maximum angles geometrically.
(Original post by Vorsah)
I think I was getting confused between polar and parametric equations. What would the point (1,0) be in parametric equations?
But polar equations are parametric equations.
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(Original post by Vorsah)
I think I was getting confused between polar and parametric equations. What would the point (1,0) be in parametric equations?
It depends on the how x and y are linked to the parameter. They can be linked in any arbitrary manner.

In the example given by you above, changing some symbols to make it clearer, x = a\cos t and y = 1+  a\sin t. This defines a circle centre cartesian(0, 1) with radius a. The point cartesian(1, 0) may or may not be on this curve, depending on the value of a. It is on the curve only when a = \sqrt{2}. This occurs at t = -\frac{\pi}{4}

A parametric curve that does definitely pass through the point is x = t^2, y = t^{\frac{1}{2}} -1 and this occurs at t = 1.

(Original post by Jai Sandhu)
But polar equations are parametric equations.
They are very similar but have subtle differences. They have the restriction that \theta is the angle between the positive x axis and the line joining the origin to the point you are concerned with.
In the first example at the top of my post, it is not polar because t is the angle between the positive x axis and the line joining the point cartesian(0, 1) to the point you are concerned with. This is not the same as the angle mentioned in my previous statement.
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