My question is concerning the distance travelled through an annulus of two concentric circles. I hope to use this to determine the source of radiation in a cylinder. The difficulty comes from the fact that I need the distance to be described as a function, the distance travelled is not directly outwards from the origin and that the inner circle has an unknown radius.
I made a diagram to illustrate the problem, I don't have enough reputation to post it so the link is here.
R and D are well known constants but r is unknown (and constant), the magnitude of r could be anywhere from 0 to R. Theta cycles from 0 to 2 pi (the little ball traces a circular orbit with a radius r). I am interested in finding an equation for L (the red line) as a function of the other 4 variables (r,R,D & theta) where theta cycles through 0 to 2 pi whilst the other three variables remain constant.
As theta increases above pi/2 part of the red line will be present in the inner circle, I would like for that part to be included in the length too.
I have tried applying the various circle theorems, cosine rule and forming congruent triangles to no avail and was hoping someone would be able to point me in the right direction.
Failing that is there a way to determine the length travelled in a straight line from a point within a circle to one outside?
EDIT: I realised that the points I'm interested in lie on the intersection of a straight line and the equation of two circles, I believe the solution may now lie in solving their simultaneous equations.
OP here. Think I was able to find a solution but I'd love to know if a more elegant solution exists.
The trick was to determine the coordinates of the intersection of the line with the outer circle. It's the intersection of a straight line: $$y_R=mx_R+c$$ with the equation for a circle: $$ R^2=y_R^2+x_R^2$$ when $y=mx+c$ is subbed into this equation I got equation 1:
$$R^2= mx_R^2 +c^2 +2mcx_R+x_R^2$$
it turns out $m$ and $c$ can be written as $$m=\frac{y_r}{D-x_r} = \frac{rsin(\theta)}{D-rcos(\theta)}$$ (looking at the triangle formed in the diagram) $$ c = Dm = \frac{Drsin(\theta)}{D-rcos(\theta)}$$
Subbing $Dm =c$ into equation 1 I rewrote the equation to group terms of $x_R$ then used the quadratic formula to determine a value for $x_R$
$$ 0 = (m+1)x_R^2 +(2m^2D)x_R+ (D^2m^2-R^2)$$
$$ x_R = \frac{(-2m^2D)\pm\sqrt{4m^4D^2-4(m+1)(D^2m^2-R^2)}}{2(m+1)}$$
It appears that two solutions exist as expected, the one I am interested is always going to be positive due to the nature of the problem.
$$ x_R = \frac{(-2m^2D)+\sqrt{4m^4D^2-4(m+1)(D^2m^2-R^2)}}{2(m+1)}$$
and in full:
$$ x_R = \frac{-2\left(\frac{rsin(\theta)}{D-rcos(\theta)}\right)^2D+\sqrt{4\left(\frac{rsin(\theta)}{D-rcos(\theta)}\right)^4D^2-4\left(\frac{rsin(\theta)}{D-rcos(\theta)}+1\right)\left(D^2\left(\frac{rsin(\theta)}{D-rcos(\theta)}\right)^2-R^2\right)}}{2\left(\frac{rsin(\theta)}{D-rcos(\theta)}+1\right)}$$
With the $x_R$ value calculated the value for $y_R$ can be calculated from the equation for a straight line.
With $x_r$, $x_R$ & $y_r$, $y_R$the length $L$ can then be calculated from simple geometry.