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Math Help - Lines tangent to a point on a circle in the complex numbers.

  1. #1
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    Lines tangent to a point on a circle in the complex numbers.

    Hello all, the question I am trying to prove is as follows:

    Let C be the circle {z: |z-c| = r},   \; r > 0;

    and let a = c+ r \; cis \; \alpha aand put

    L_{\beta}= \left\{ z: Im(\frac{z-a}{b})=0\right\}

    where b = cis \; \beta

    Find necessary and sufficient conditions in terms of \beta that L_{\beta} be tangent to C at a

    Now if L_{\beta} was tangent to the circle  C at the point a, L_{\beta} and CA (i.e. the line segment from c to a, which has distance r) would have to be perpendicular.

    i.e. triangle ACO, (where O is any point z= a + tb where (-\infty < t < \infty) on L_{\beta} ) is a right triangle.

    However using the Pythagorem theorem on the sides, doesnt seem to help me find what \beta is.

    Can anyone offer me any advice on this problem please?
    Last edited by Sheld; March 11th 2011 at 10:46 AM. Reason: llatex formatting.
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  2. #2
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    Quote Originally Posted by Sheld View Post
    Hello all, the question I am trying to prove is as follows:

    Let C be the circle {z: |z-c| = r},   \; r > 0;

    and let a = c+ r \; cis \; \alpha aand put

    L_{\beta}= \left\{ z: Im(\frac{z-a}{b})=0\right\}

    where b = cis \; \beta

    Find necessary and sufficient conditions in terms of \beta that L_{\beta} be tangent to C at a

    Now if L_{\beta} was tangent to the circle  C at the point a, L_{\beta} and CA (i.e. the line segment from c to a, which has distance r) would have to be perpendicular.

    i.e. triangle ACO, (where O is any point z= a + tb where (-\infty < t < \infty) on L_{\beta} ) is a right triangle.

    However using the Pythagorem theorem on the sides, doesnt seem to help me find what \beta is.

    Can anyone offer me any advice on this problem please?


    From analytic geometry: two non-vertical lines are perpendicular iff the product of their slopes is -1...Now,

    \displaystyle{z=x+iy\in L_\beta\Longleftrightarrow \tan\beta=\frac{Im\,z-Im\, c}{Re\,z-Re\,c} , and the slope of ca is m_{ac}=\tan\alpha ...

    Tonio
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  3. #3
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    Oh ok, I think I understand so

    tan(\beta) = \frac{-1}{tan(\alpha)}\;\;\;\; ?

    Sorry if I am not understanding. Is tan(\beta) = \frac{Im \; z - Im \;c}{Re \; z - Re \;c} the slope of L_{\beta} or the line from the center c to a point on L_{\beta}?

    Thank you very much for the reply.
    Last edited by Sheld; March 11th 2011 at 04:16 PM. Reason: latex formatting
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