# Question relating to shifting a circle.

1. Jul 2, 2012

### ozone

This problem arose for me while working out a triple integral in spherical coordinates. Basicly 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?

2. Jul 2, 2012

### Staff: Mentor

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

3. Jul 2, 2012

### SammyS

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

Here's a ploy from WolframAlpha:

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4. Jul 2, 2012

### ozone

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

5. Jul 2, 2012

### SammyS

Staff Emeritus
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.