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

1. ## 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.

2. Originally Posted by Sheld
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.

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

3. 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.