Question

Two different nonparallel lines cut the circle $\left|z\right|=r$ at points a, b, c, and d, respectively. Then show that these lines meet at the point given by
$\frac{a^{-1}+b^{-1}-c^{-1}-d^{-1}}{a^{-1}{b}^{-1}-c^{-1}{d}^{-1}}$ .

Answer 1

Since, P, Q, R are collinear, we have
$\left|\begin{array}{ccc}z& \overline{z}& 1\\ c& \overline{c}& 1\\ d& \overline{d}& 1\end{array}\right|=0$
$⟹z(\overline{c}-\overline{d})-\overline{z}(c-d)+(c\overline{d}-c\overline{d})=0$ (1)
Similarly,
$z(\overline{a}-\overline{b})-z(a-b)+(a\overline{b}-a\overline{b})=0$ (2)
From $\left\{\right(1)\times (a-b\left)\right\}-\left\{\right(2)\times (c-d\left)\right\}$, we get
$z\left[\right(\overline{c}-\overline{d}\left)\right)(a-b)-(\overline{a}-\overline{b})(c-d)]$
$=(a\overline{b}-b\overline{a})(c-d)-(c\overline{d}-\overline{c}d)(a-b)$ (3)
Now,
${\left|a\right|}^2=a\overline{a}=r^2$ or $\overline{a}=\frac{r^2}{a}$
Similarly,
$\overline{b}=\frac{r^2}{b},\overline{c}=\frac{r^2}{c},\overline{d}=\frac{r^2}{d}$
From (3), we get
$z\left[(\frac{r^2}{c}-\frac{r^2}{d})\right(a-b)-(\frac{r^2}{a}-\frac{r^2}{b})(c-d\left)\right]$
$=(\frac{a{r}^2}{b}-\frac{b{r}^2}{a})(c-d)-(\frac{c{r}^2}{d}-\frac{d{r}^2}{c})(a-b)$
or $z[-\frac{1}{cd}+\frac{1}{ab}]=\frac{(a+b)}{ab}-\frac{c+d}{cd}$
or $z=\frac{a^{-1}+b^{-1}-c^{-1}-d^{-1}}{a^{-1}{b}^{-1}-c^{-1}{d}^{-1}}$
Ans: 1