A question about simple circles and triangles

Question

Today I've encountered a question like the following I am adding a picture because I have to;

The question paragraph says;

$\text{Given} \quad |OF|=6 \quad \text{and} \quad |BF|=4$

What is $|CH|=x$

My Attempts;

I have noticed that the diameter $r=10$ (1)

I have drawn a line from $C$ to $O$ which also is $r$ (2)

I have written $|HO|=\sqrt{100-x^2}$ but couldn't go further,

What do you suggest?

Answer 1

Let $\alpha=\angle HFC$ and $a=EF$, so that $x=2a\sin\alpha$.

From the cosine law applied to triangle $OFC$ we get: $$ OF^2+FC^2-2OF\cdot FC\cos\alpha=OC^2, \quad\hbox{that is:}\quad a^2-6a\cos\alpha=16. $$

From the cosine law applied to triangle $OFD$ (notice that $FD=\sqrt2a$) we get: $$ OF^2+FD^2-2OF\cdot FD\cos(\pi/4+\alpha)=OD^2, \quad\hbox{that is:}\quad a^2-6a\cos\alpha+6a\sin\alpha=32. $$ Substituting here our previous result we thus get $6a\sin\alpha=16$, that is $x=2a\sin\alpha={16\over3}$.

Answer 2

This is only a partial solution, because I get the results using analytic geometry

Anyway I hope it can be useful