Question

Two different non parallel lines cut the circle $\left|z\right|=r$ in point $a,b,c,d$ respectively. Prove that these lines meet in the point $z$ given by $z=\frac{a^{-1}+b^{-1}-c^{-1}-d^{-1}}{a^{-1}{b}^{-1}-c^{-1}{d}^{-1}}$

Answer 1

Since point $P,A,B$ are collinear

$\therefore$ $\left|\begin{array}{ccc}z& \overline{z}& 1\\ a& \overline{a}& 1\\ b& \overline{b}& 1\end{array}\right|=$

$\Rightarrow$ $z(\overline{a}-\overline{b})-\overline{z}(a-b)+(a\overline{b}-\overline{a}b)=0$ (i)

Similarly, points $P,C,D$ are collinear, so

$z(\overline{c}-\overline{d})-\overline{z}(c-d)+(c\overline{d}-\overline{c}d)=0$

On applying $\left(i\right)\times (c-d)-\left(ii\right)(a-b),$ we get

$\therefore$ $z(\overline{a}-\overline{b})(c-d)-z(\overline{c}-\overline{d})(a-b)=(c\overline{d}-\overline{c}d)(a-b)-(a\overline{b}-\overline{a}b)(a-d)$ (iii)

$\because$ $z\overline{z}=r^2=k$ (say)

$\therefore$ $\overline{a}=\frac{k}{a},\overline{b}=\frac{k}{b},\overline{c}=\frac{k}{c}$ etc.

From equation (iii) we get

$z(\frac{k}{a}-\frac{k}{b})(c-d)-z(\frac{k}{c}-\frac{k}{d})(a-b)=(\frac{ck}{d}-\frac{kd}{c})(a-b)-(\frac{ak}{b}-\frac{bk}{a})(c-d)$

$\therefore$ $z=\frac{a^{-1}+b^{-1}-c^{-1}-d^{-1}}{a^{-1}{b}^{-1}-c^{-1}{d}^{-1}}$