What is the length of the segment at angle $\alpha$ from a point on the diameter to the circle?

Question

In Proposition III.7 (see also modern version), Euclid shows that in the diagram below $FA > FB > FC > FG > FE$.

Assuming $\angle CFE = \alpha$, what is the actual length of $FC$ (in terms of $FD$) on a unit circle? I tried to determine this algebraically but was not able to (some attempts are here, here, and here, abbreviating $u = FD$).

I also tried to solve geometrically using Power of the Point (again, letting $u = FD$ and $C'$ be the other intersection of $FC$ with the circle): $$(1+u)(1-u) = FC \cdot FC' \\ FC + FC' = ? $$ but could not express $CC'$ in terms of $\alpha$ (or anything else useful).

Answer 1

Letting $u=DF, v=CF, \alpha = \angle CFE, \alpha' = \angle CFD, \beta = \angle CDF, \gamma = \angle DCF$, and noting that $CD = 1$ because it is a radius of the circle, we can use the law of sines to set up two equations of two unknowns:

$$\frac{\sin{\alpha'}}{1} = \frac{\sin{\beta}}{v} = \frac{\sin{\gamma}}{u}$$

Note that $\beta = \alpha - \gamma$ so:

$$\sin{\alpha'} = \frac{\sin{\alpha - \gamma}}{v} = \frac{\sin{\gamma}}{u}$$

Thus $\gamma = \arcsin{(u\sin{\alpha'})}$. Recognize that $\sin{(\alpha - \gamma)} = \sin{\alpha}\cos{\gamma} - \sin{\gamma}\cos{\alpha}$ so:

$$\sin{\alpha'} = \frac{\sin{\alpha}\cos{\gamma} - \cos{\alpha}\sin{\gamma}}{v}$$

Next we realize that $\cos{\gamma} = \sqrt{1 - u^2\sin{\alpha'}}$ and $\sin{\gamma} = u\sin{\alpha'}$ so we end up with:

$$\sin{\alpha'} = \frac{\sin{\alpha}\sqrt{1 - u^2\sin^2{\alpha'}} - u\cos{\alpha}\sin{\alpha'}}{v}$$

Which, solving for $v$ and simplifying, gives us:

$$v = CF = \sin{\alpha}\sqrt{\frac{1}{\sin^2{\alpha'}} - u^2}-u\cos{\alpha}$$

The following plot compares the simulated distance CF as well as the calculated distance using the derived formula (setting $u=0.2$).

Plot of calculated CF distance and simulated distance against $\alpha$

The following Python code produced that plot

import numpy as np
import matplotlib.pyplot as plt

alpha = np.linspace(1e-5, np.pi-1e-5, 1000)
u = 0.2
point_F = np.array([u, 0])
point_C = np.array([np.array([1, 0])*np.cos(a) + np.array([0, 1])*np.sin(a) for a in alpha])
CF_simulated = np.array([np.linalg.norm(p_c - point_F) for p_c in point_C])
CF_calculated = np.sin(alpha)*np.sqrt(1/np.sin(np.pi - alpha)**2 - u**2) - u*np.cos(alpha)
plt.plot(alpha, CF_simulated)
plt.plot(alpha, CF_calculated)
plt.legend(['CF_simulated', 'CF_calculated'])
plt.xlabel(r'$\alpha$ (rad)')
plt.ylabel('distance CF')
plt.show()