- #1

Icebreaker

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, we discussed a circle with a radius R centered at the origin and a positive y-intercept at (0, R). As the circle is shifted by n on the x-axis, the y-intercept decreases. We wondered if we could determine the radius given a certain proportion between the x-shift and the decrease in y-intercept. We then considered the equation x^2 + y^2 = (x + n)^2 + (y + m)^2 and the degenerative case x = 0. We also discussed a formula for the radius R, simplified from the equation (y - 5)^2 + 45^2 = y^2, as R=(a^2+b^2)/2a,

- #1

Icebreaker

Mathematics news on Phys.org

- #2

honestrosewater

Gold Member

- 2,143

- 6

I don't know what to do about the degenerative case, but could you solve x^{2} + y^{2} = (x + n)^{2} + (y + m)^{2} given only n and m?

Edit: Maybe I do know what to do in the degenerative case- x = 0. So your example is y^{2} = 45^{2} + (y - 5)^{2}

Edit: Maybe I do know what to do in the degenerative case- x = 0. So your example is y

Last edited:

- #3

Icebreaker

- #4

BicycleTree

- 520

- 0

http://img195.echo.cx/my.php?image=circle9qu.png

So, (y - 5)^2 + 45^2 = y^2 as honestrosewater said.

- #5

honestrosewater

Gold Member

- 2,143

- 6

I don't how many solutions there are for y^{2} = 45^{2} + (y - 5)^{2}. I was just considering that the circle's center C, the graph's origin O, and the y-intercept Y form a right triangle with line CY as the hypotenuse & radius, and the radius is constant.

Edit: But consider that eventually CY = CO...

Edit: But consider that eventually CY = CO...

Last edited:

- #6

Icebreaker

- #7

Kamataat

- 137

- 0

[tex]R=\frac{a^2+b^2}{2a}[/tex]

where "a" is the change in y-intercept and "b" is the shift in the x direction.

- Kamataat

- #8

BicycleTree

- 520

- 0

- #9

Kamataat

- 137

- 0

- Kamataat

- #10

Icebreaker

Indeed, a general formula for a situation like this may very well be

[tex](r+\Delta y)^2 + (\Delta x)^2 = r^2, \Delta x < r/2[/tex]

I'm not sure I can prove it, or whether it has already been proven, or whether it's trivial.

[tex](r+\Delta y)^2 + (\Delta x)^2 = r^2, \Delta x < r/2[/tex]

I'm not sure I can prove it, or whether it has already been proven, or whether it's trivial.

Last edited by a moderator:

The equation of a circle of radius R centered at the origin is **x ^{2} + y^{2} = R^{2}**.

The area of a circle can be found using the formula **Area = πR ^{2}**, where R is the radius of the circle.

The circumference of a circle can be found using the formula **Circumference = 2πR**, where R is the radius of the circle.

To graph a circle of radius R centered at the origin, plot the point (0,0) as the center of the circle and then plot points along the circle's circumference using the equation **x ^{2} + y^{2} = R^{2}**.

The diameter of a circle is always twice the length of the radius. So, for a circle of radius R centered at the origin, the diameter is equal to 2R.

- Replies
- 2

- Views
- 965

- Replies
- 2

- Views
- 1K

- Replies
- 4

- Views
- 986

- Replies
- 6

- Views
- 1K

- Replies
- 9

- Views
- 2K

- Replies
- 4

- Views
- 1K

- Replies
- 2

- Views
- 1K

- Replies
- 5

- Views
- 2K

- Replies
- 8

- Views
- 970

- Replies
- 4

- Views
- 2K

Share: