Question

Two different non parallel lines meet the circle $|z|=r$. One of them at points a and b and the other which is tangent to the circle at c.

Then show that the point of intersection of two lines is $\frac{2c^{-1}-a^{-1}-b^{-1}}{c^{-2}-a^{-1}{b}^{-1}}$.

where a, b, c are complex numbers.

Answer 1

Let two non parallel straight lines PQ, RS meet the circle $|z|=r$ in the points $a,bandc$
then $|a|=r,|b|=r$ and $|c|=r$
or $|a{|}^2=|b{|}^2=|c{|}^2=r^2$
$\therefore a\overline{a}=b\overline{b}=c\overline{c}=r^2$
then $\overline{a}=\frac{r^2}{a},\overline{b}=\frac{r^2}{b}$ and $\overline{c}=\frac{r^2}{c}$
Points $a,b,z$ are collinear
then $\left|\begin{array}{ccc}z& \overline{z}& 1\\ a& \overline{a}& 1\\ b& \overline{b}& 1\end{array}\right|=0$
$\therefore z(\overline{a}-\overline{b})-\overline{z}(a-b)+a\overline{b}-\overline{a}b=0$
$\Rightarrow z(\frac{r^2}{a}-\frac{r^2}{b})-\overline{z}(a-b)+\frac{r^{2}a}{b}-\frac{r^{2}b}{a}=0$
Dividing both sides by $r^2(b-1)$
$\therefore \frac{z}{ab}+\frac{\overline{z}}{r^2}=a^{-1}+b^{-1}$ ........ (1)
If $O$ is the centre of the circle then
$\Rightarrow |z{|}^2=|z-c{|}^2+|c-0{|}^2$
$\Rightarrow z\overline{z}=(z-c)(\overline{z}-\overline{c})+c\overline{c}$
$\Rightarrow z\overline{z}=z\overline{z}-z\overline{c}-c\overline{z}+c\overline{c}+c\overline{c}$
$\therefore z\overline{c}+\overline{z}c=2c\overline{c}$
$\Rightarrow \frac{z{r}^2}{c}+\overline{z}c=\frac{2c{r}^2}{c}$
$\Rightarrow \frac{z{r}^2}{c}+\overline{z}c=2r^2$
Dividing both sides by $c{r}^2$
$\therefore \frac{z}{c^2}+\frac{\overline{a}}{r^2}=2c^{-1}$ ........... (2)
Subtracting (2) and (1)
We have $z(c^{-1}-a^{-1}{b}^{-1})=2c^{-1}-a^{-1}-b^{-1}$
$\therefore z=\frac{2c^{-1}-a^{-1}-b^{-1}}{c^{-2}-a^{-1}{b}^{-1}}$