Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $AX$.

share|cite|improve this question
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


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 $\,3\,$ .

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.