If I have a line $AB$ intersecting a circle at an angle of $\theta$1 from the center of the circle and change it by an amount of $\theta$2 so it intersects at point $C$ instead, how long will the arc $BC$ be and how much longer will the line $AC$ be?
Answers in terms of either $R$ or $D1$ would be equally useful to me if it's easier to express in terms of one than the other.
It is my deepest wish to post an image alongside this explanation. :-( I'll try to explain it in words. This comes from your excellent image: Let $D_1$ be the length of ${AB}$, $D_2$ be the length of ${AC}$, $\theta_1$ be the angle made by $AB$ and the $x$-axis, and $\theta_2$ be the angle made by $AC$ and the $y$-axis.
New stuffs: So, draw two lines from the center of the circle to points $B$ and $C$. These have length $R$. Let $O$ represent the center of the circle. The angle $OB$ makes with the $x$-axis is $\alpha$, and $OC$ likewise is $\beta$. Finally, $M$ is the intersection of the circle and the positive $x$-axis.
Everything that follows assumes that you know the values of $D_1$ and $D_2$.
If you don't, then you need to know the length of $AM$ and $BM$ and $CM$, and then use some really fancy triangle properties. Life marches on, though! (If something's wrong, comment it and I'll work with it. Glad to help!)
Using the Law of Sines, we can determine that ${\sin(\alpha)\over D_1}={\sin(\theta_1) \over R}$ --> $\alpha = \sin^{-1}\left(\frac{D_1\sin(\theta_1)}R\right)$
Similarly, $\alpha + \beta = \sin^{-1}\left(\frac{D_2\sin(\theta_1+\theta_2)}R\right)$
Length of an arc is radians times radius.
Arc $\overparen{BC} = R((\alpha + \beta) - \alpha) = R\left(\sin^{-1}\left(\frac{D_1\sin(\theta_1)}R\right) - \sin^{-1}\left(\frac{D_2\sin(\theta_1+\theta_2)}R\right)\right)$.
Reading over the question again, I just realized you don't know what $D_1$ and $D_2$ are. I will work on those and be back by tomorrow. Have fun! Hope this helps!