Internal point in circle and chords

Question

I'm stucked in this problem, and I can't understand what I'm missing.

This is the problem, we have a point $C$ in the circle, not in the centre, connected to the center with the segment $DE$, this segment is fixed. Then we have an angle in $C$, $ECF$, that meets the circle in $F$. In the picture I've also connected $C$ and $F$ to find the segment $FG$.

Now the question is, given $CE=H$, $AE=AD=R$, where A is the center of the circle and R the radius, and given the angle $\theta$, can we find the length of $CF$?

Until now I've found from a theorem that $EC \cdot CD=GC \cdot CF$, or that $H(2R-H)=xy$, if we call $CF=x$ and $GC=y$.

I think I have to find some other relation between the segments, but I don't find anything useful.

Answer 1

By analytical geometry:

The parametric equations of the line $CF$ are

$$\begin{cases}x=t\sin\theta,\\y=H-R-t\cos\theta\end{cases}$$ and you intersect the circle

$$x^2+y^2=R^2.$$

Hence,

$$t^2-2(H-R)\cos\theta\,t+H^2-2HR=0$$ and

$$t=(H-R)\cos\theta\pm\sqrt{(H-R)^2\cos^2\theta-H^2+2HR}.$$

The requested length is the positive solution.

Answer 2

Using high school geometry. Consider isosceles $\triangle{AGF}$. Let $AX$ be the height. Then $AX=(R-CE)\sin \theta$. (This is from $\triangle{ACX}$). Now $\frac{1}{2} FG=\sqrt{R^2-AX^2}$

To find $CF$, let's find $\sin \angle{AFG}=\frac{R}{AX}$. Now we can find $\angle{CAF}=\theta-\arcsin{\frac{R}{AX}}$.

Finally, using law of cosines, $$CF^2=(R-H)^2+R^2-2(R-H)R \cos \angle{CAF}$$