# Dissected Circle Formula Help

• Jan 14th 2008, 11:35 AM
Ken010101
Dissected Circle Formula Help
Problem:
Find the distance, from the circle center to a line, where the line dissects the circle into areas of 3/4 and 1/4.

Any help with a formula, such that I can calculate this distance for various circle diameters would be greatly appreciated. I could add a pictures, if I have not described my problem clearly.

Regards,
Ken
• Jan 14th 2008, 11:56 AM
earboth
Quote:

Originally Posted by Ken010101
Problem:
Find the distance, from the circle center to a line, where the line dissects the circle into areas of 3/4 and 1/4.

...

Hello,

1. I've attached a drawing.

2. Using Pythagorean theorem you can derive a set of useful formula:

$\displaystyle h = r - \frac12 \cdot \sqrt{4r^2-s^2}$

$\displaystyle s = 2\sqrt{2hr-h^2}$

$\displaystyle A_{segment} = A_{sector} - A_{triangle}$

If you are allowed to use trigonometric functions the area of the segment can be calculated by:

$\displaystyle A_{segment}=\frac{r^2}{2}\cdot \left(\frac{\pi \alpha}{180^\circ}-\sin(\alpha)\right)$
• Jan 14th 2008, 01:03 PM
Ken010101
Thank you very much earboth, for the quick reply.
Quote:

Originally Posted by earboth
Using Pythagorean theorem you can derive a set of useful formula:

$\displaystyle h = r - \frac12 \cdot \sqrt{4r^2-s^2}$

$\displaystyle s = 2\sqrt{2hr-h^2}$

So, for me to solve for h, am I understanding this correctly?
$\displaystyle h = r - \frac12 \cdot \sqrt{4r^2-(2\sqrt{2hr-h^2})^2}$

Is it possible for you to further assist me in simplifying this, so I can solve for h? Sorry, but I just don't understand where to begin...

I cannot tell you how appreciative I am.
Ken
• Jan 14th 2008, 01:52 PM
topsquark
Quote:

Originally Posted by Ken010101
Thank you very much earboth, for the quick reply.

So, for me to solve for h, am I understanding this correctly?
$\displaystyle h = r - \frac12 \cdot \sqrt{4r^2-(2\sqrt{2hr-h^2})^2}$

Is it possible for you to further assist me in simplifying this, so I can solve for h? Sorry, but I just don't understand where to begin...

I cannot tell you how appreciative I am.
Ken

$\displaystyle h = r - \frac{1}{2} \cdot \sqrt{4r^2 - 4(2hr - h^2)}$

$\displaystyle h - r + \frac{1}{2} = \sqrt{4r^2 - 4(2hr - h^2)}$

$\displaystyle \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)$

(I made minor mistake after this point, which would be too long to correct. Please refer to earboth's post below for some of the next steps.)

-Dan
• Jan 14th 2008, 09:04 PM
earboth
Quote:

Originally Posted by topsquark
$\displaystyle h = r - \frac{1}{2} \cdot \sqrt{4r^2 - 4(2hr - h^2)}$

$\displaystyle h - r + \frac{1}{2} = \sqrt{4r^2 - 4(2hr - h^2)}$

$\displaystyle \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)$

$\displaystyle h^2 + r^2 + \frac{1}{4} - 2hr + \frac{h}{2} - r = 4r^2 - 8hr - 4h^2$

......

-Dan

Good morning,

I don't want to pick at you but you have made a minor mistake ....

$\displaystyle \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)$

$\displaystyle h^2+r^2+\frac14+h-r-2hr=4r^2-8hr+4h^2$

$\displaystyle 4h^2-8hr+4h+4r^2-4r=16h^2-32hr+16r^2-1$

After a few steps you'll get:

$\displaystyle h=\frac{6r-1}{6}~\vee~h=\frac{2r+1}{2}$

EDIT: In my opinion this result is much too simple so there must be a mistake in the original equation
• Jan 14th 2008, 11:52 PM
earboth
Quote:

Originally Posted by Ken010101
Problem:
Find the distance, from the circle center to a line, where the line dissects the circle into areas of 3/4 and 1/4.

Any help with a formula, such that I can calculate this distance for various circle diameters would be greatly appreciated. I could add a pictures, if I have not described my problem clearly.

Regards,
Ken

Hello,

all those nice formulae I gave you lead to monstrous equation which I cann't handle anymore.

So I decided to go back to the roots: I considered a half-circle to be the graph of a function

$\displaystyle c(x) = \sqrt{r^2-x^2}$

Then I used integration

$\displaystyle \int_u^r \left(\sqrt{r^2-x^2} \right)dx = \frac18 \cdot \pi \cdot r^2$

Then the height of the segment must be $\displaystyle h = r-u$

I've attached a screenshot of the calculations (which were made by my computer)
• Jan 15th 2008, 06:58 AM
topsquark
Quote:

Originally Posted by earboth
Good morning,

I don't want to pick at you but you have made a minor mistake ....

$\displaystyle \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)$

$\displaystyle h^2+r^2+\frac14+h-r-2hr=4r^2-8hr+4h^2$

$\displaystyle 4h^2-8hr+4h+4r^2-4r=16h^2-32hr+16r^2-1$

After a few steps you'll get:

$\displaystyle h=\frac{6r-1}{6}~\vee~h=\frac{2r+1}{2}$

EDIT: In my opinion this result is much too simple so there must be a mistake in the original equation

Please, go ahead and "pick" at me! I'm always glad to have someone check my work. And good spot. (Clapping)

-Dan
• Jan 15th 2008, 09:35 PM
earboth
Quote:

Originally Posted by topsquark
$\displaystyle h = r - \frac{1}{2} \cdot \sqrt{4r^2 - 4(2hr - h^2)}$

$\displaystyle h - r \boxed{+ \frac{1}{2}} = \sqrt{4r^2 - 4(2hr - h^2)}$

$\displaystyle \left ( h - r + \frac{1}{2} \right ) ^2 = 4r^2 - 4(2hr - h^2)$

(I made minor mistake after this point, which would be too long to correct. Please refer to earboth's post below for some of the next steps.)

-Dan

Hello,

as you have certainly spotted yourself I was blinded a little bit. The mm occured some steps earlier. $\displaystyle \boxed{\text{I have boxed the algebraic accident}}$