Question:-
Three points represented by the complex numbers $a,b$ and $c$ lie on a circle with center $O$ and radius $r$. The tangent at $c$ cuts the chord joining the points $a$ and $b$ at $z$. Show that $$z=\dfrac{a^{-1}+b^{-1}-2c^{-1}}{a^{-1}b^{-1}-c^{-2}}$$
Attempt at a solution:- To simplify our problem let $O$ be the origin, then the equation of circle becomes $|z|=r$.
Now, the equation of chord passing through $a$ and $b$ can be given by the following determinant
$$\begin{vmatrix}
z & \overline{z} & 1 \\
a & \overline{a} & 1 \\
b & \overline{b} & 1 \\
\end{vmatrix}= 0$$
which simplifies to $$z(\overline{a}-\overline{b})-\overline{z}(a-b)+(\overline{a}b-a\overline{b})=0 \tag{1}$$
Now, for the equation of the tangent through $c$, I used the cartesian equation of tangent to a circle $xx_1+yy_1=r^2$ from which I got $$z\overline{c}+\overline{z}c=2r^2\tag{2}$$
Now, from equation $(1)$, we get
$$\overline{z}=\dfrac{z\left(\overline{a}-\overline{b}\right)+\left(a\overline{b}-\overline{a}b\right)}{(a-b)}$$
Putting this in equation $(2)$, we get $$z=\dfrac{2r^2(a-b)+\left(a\overline{b}-\overline{a}b\right)c}{\left(a\overline{c}+\overline{a}c\right)-\left(b\overline{c}+\overline{b}c\right)}$$
After this I am not able to get to anything of much value, so your help would be appreciated. And as always, more solutions are welcomed.
Best Answer
Denoting a vector as difference of position vectors,so
angles $ bcz = cab,$ so
$$ bc \cdot cz / ( |bc|\, |cz| ) = ca \cdot ab / ( |ca|\, |ab| ) $$