Question relating to shifting a circle.

[ozone]

This problem arose for me while working out a triple integral in spherical coordinates. basically I know that when we shift a parabola along the axis it is simply translated. I naturally assumed that if we shifted a circle in a similar manner that it would act the same.

However when we shift a circle along the axis, such as one with the equation

$(x-1)^2 + y^2 = 1$ We find that the entirety of the circle now sits above the x axis, and that our radius becomes $2cos∂$

Could anyone shed some light on this?

 

[jedishrfu]

you've moved the center to (1,0) but the circle should still have a radius of 1

try some points (0,0) , (1,1) , (1,1) and (1,-1) all satisfy the equation and show that the radius is 1 and not as you say and that ithe circle still lies on the x-axis

 

[SammyS]

This problem arose for me while working out a triple integral in spherical coordinates. Basically I know that when we shift a parabola along the axis it is simply translated. I naturally assumed that if we shifted a circle in a similar manner that it would act the same.

However when we shift a circle along the axis, such as one with the equation

$(x-1)^2 + y^2 = 1$ We find that the entirety of the circle now sits above the x axis, and that our radius becomes $2cos∂$

Could anyone shed some light on this?

$(x-1)^2 + y^2 = 1$ is the equation of a circle having radius 1 with its center at (1, 0).

Here's a ploy from WolframAlpha:

 

[ozone]

Alright I guess I just misunderstood the solution on the problem set.. thank you.

 

[SammyS]

... We find that the entirety of the circle now sits above the x axis, and that our radius becomes $2\cos\theta$

Could anyone shed some light on this?

After reading your Original Post, it looks as if you might be converting $(x-1)^2 + y^2 = 1$ to polar coordinates with the result, $r=2\cos\theta\,.$ If so, that variable, r, does not refer to the radius of the circle, it's the the distance that the point (x, y) is from the origin.