If you have two intersecting circles with different length radii as shown here (and adapted/borrowed from another question), is the ratio of the two intersected arc's sagitta (FE and GE) equivalent to the ratio of their radii (AG and CF)?
If not, what is the best way to determine the lengths of the individual sagitta if only the total amount of overlap (FG) and the circles' radii are known?
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In unequal circles with a common chord, the circle with the lesser radius has the greater sagitta, since the "bow" is more bent. Therefore the sagittae are not directly proportional to the radii. Suppose they are inversely proportional. In the posted figure, extending GA and FC to meet the circles at H and K, and joining BH, BF, BG, BK, then since $$\frac{EG}{EB}=\frac{EB}{EH}$$ we have $$EG = \frac{(BE)^2}{EH}$$ Similarly we get $$EF = \frac{(BE)^2}{EK}$$ Therefore $$\frac{EG}{EF}=\frac{EK}{EH}=\frac{R_2}{R_1}$$ But suppose the overlap is such that BD is the diameter of the lesser circle. Then since $$EK=R_2$$ it follows that $$EH=R_1$$ which is false, since BD is less than the diameter of the greater circle. Therefore the sagittae are not inversely proportional to the radii.
No, the ratio is not, in general, the same. If $L$ is the length of $BE$, and $R_1$ is the length of $AB$ then $$ EG = R_1\left(1 - \sqrt{1-\left(\frac{L}{R_1}\right)^2}\right) $$ while $$ EF = R_2\left(1 - \sqrt{1- \left(\frac{L}{R_2}\right)^2}\right) $$ so that the ratio is \begin{align} \rho &= \frac{R_1}{R_2} \frac{1 - \sqrt{1-\left(\frac{L}{R_1}\right)^2}} {1 - \sqrt{1-\left(\frac{L}{R_2}\right)^2}} \end{align} which differs from the ratio of the radii by a factor of \begin{align} s &= \frac{1 - \sqrt{1-\left(\frac{L}{R_1}\right)^2}} {1 - \sqrt{1-\left(\frac{L}{R_2}\right)^2}}\\ \end{align}
I got the formula for the sagitta from wikipedia.