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 $AX$.

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Since the tangent at point $B$ lies outside the circle, when you say, "Tangent of the circle $w$ on the point $B$ intersect $CE$ in the point X," do you mean the line passing through point $CE$? By your description, the point $X$ lies outside the circle $w$, and hence could not intersect the line segment $CE$. Do I understand you correctly? – Carl Morris Sep 21 '12 at 18:39

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 $\,3\,$ .

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