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Given a point A that outside a circle so that $AT$ is tangent to the circle in point $T$

And $AC$ is a secant to that circle in points $B,C$.

From points $B,C$ we build heights to $AT$ in points $D,E$. (point $E$ is between points $D,T$)

$BD=b$, $CE=c$, angle $TAC=\alpha$.

What's the expression of radius $r$ in terms of $b,c,\alpha$?

I'm trying to find a nice solution maybe euclidean or trigonometric/vectors.

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Welcome to math.SE! What did you already try to solve it? Besides, what do you know about it? Without knowing the tools you have at your disposal, it is impossible to explain anything. –  mau Oct 31 '13 at 10:12
    
thanks! i tried the basics like law of sine, cosine, similarity. –  dob Oct 31 '13 at 10:19

1 Answer 1

up vote 3 down vote accepted

enter image description here

$AB=b\csc\alpha$, $~AC=c\csc\alpha$ $\Longrightarrow$ $AB\cdot AC=AT^{2}$, $~AT=\sqrt{bc}\csc\alpha$
$\angle BTA=\theta=\angle BCT$ $\Longrightarrow$ $2r=\dfrac{BT}{\sin\theta}=\dfrac{BT^{2}}{BD}=\dfrac{BD^{2}+DT^{2}}{BD}=\dfrac{b^{2}+(AT-AD)^{2}}{b}$
$=\dfrac{b^{2}+(\sqrt{bc}\csc\alpha-b\cot\alpha)^{2}}{b}$

$\therefore$ $r=\dfrac{b^{2}+(\sqrt{bc}\csc\alpha-b\cot\alpha)^{2}}{2b}$

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if $E$ is located between somewhere of points $D$ and $T$ (the secant is higher) will it change the result? i just can't post a sketch.. –  dob Oct 31 '13 at 12:46
    
$c~$ is used just only for $AT$'s length. so the result doesn't change. –  chloe_shi Oct 31 '13 at 12:57
1  
i see, nice solution thanks –  dob Oct 31 '13 at 13:09
    
chole_shi you are nocturnal.am i right???????????? –  krishan Oct 31 '13 at 17:45
    
I'm in South-Korea near to China. –  chloe_shi Oct 31 '13 at 17:51

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