# Thread: Circle segment of exact length

1. ## Circle segment of exact length

I need to find:

• a spherical line
• that passes through the points 0,0 and 8,8
• and the distance of the line between those two points must be exactly 12

I imagine the answer will be in the form of:
A circle centered at X,Y with a radius of R.
I'm a math newb and I don't know how to get started on the problem. Can anyone point me in the right direction? Thanks!

2. ## Re: Circle segment of exact length

What do you mean the distance between the points must be 12? The distance between (0,0) and (8,8) is $8\sqrt2$.

If you mean that the radius must be 12 then let the centre point of the circle be (a,b), using the distance formula get 2 equations for a and b.

3. ## Re: Circle segment of exact length

I imagine Simon means the distance as measured along the curve. But I am puzzled by the phrase "spherical line" when he later talks about a two dimensional circle rather than a three dimensional sphere. Simon, if you really want a circle, of some radius R, that passes through (0, 0) and (8, 8) and such that the (minor) arc of the circle with endpoints at (0, 0) and (8, 8) you can do the following:

As Shakarri says, the straight line distance between (0, 0) and (8, 8) is $8\sqrt{2}$. Using the cosine law, if the angle is $\theta$ then $128= 2R^2(1- cos(\theta))$. The arc in a circle of radius R with with central angle $\theta$ is $\frac{\theta}{2\pi}R= 12$. That gives you two equations to solve for R and $\theta$.

4. ## Re: Circle segment of exact length

Thanks, HallsofIvy. I think I've got some answers: 8*sqrt(2)=r*2*sin(a/2), 12=a/180*pi*r - Wolfram|Alpha