Finding the length of $AX$ If we assume that $AC$ is the diameter for the circle $w$ ,and the radius of this circle  is $1$ .If we assume that $D$ is a point on $AC$ such that $CD=\frac{1}{5}$.And if we assume that $B$ is a point on the circle $w$ such that $BD\perp AC$.Also $E$ is a midpoint for $BD$ .The tangent of the circle $w$ on the point $B$ intersect $CE$ in the point $X$.How to find  the length of $AX$.

With analytic geometry: center your circle at the origin $\,O\,$ in the xy-plane, so that we have the circle $\,x^2+y^2=1\,$. From the given data, we get
$$D=\left(\frac{4}{5},0\right)\,\,,\,\,B=\left(\frac{4}{5},\frac{3}{5}\right)\,\,,\,\,E=\left(\frac{4}{5},\frac{3}{10}\right)\,\,,\,\,C=(1,0)\,\,,\,\,A=(-1,0)$$
Now, let us calculate the equations of some lines:
$$m_{CE}=\frac{-3/10}{1/5}=-\frac{3}{2}\Longrightarrow CE\;:\;y=-\frac{3}{2}x+\frac{3}{2}$$
$$m_{BX}=-\frac{1}{OB}=-\frac{4}{3}\Longrightarrow BX\;:\;y=-\frac{4}{3}x+\frac{5}{3}$$
Now we can calculate the intersection point of $\,CE\,\,,\,\,BX\,$:
$$-\frac{3}{2}x+\frac{3}{2}=-\frac{4}{3}x+\frac{5}{3}\Longrightarrow x=-1\Longrightarrow y= 3\Longrightarrow X=(-1,3)$$
Thus, the distance between $\,A\,$ and $\,X\,$ is $\,3\,$ .
