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.


2 Answers 2



Denoting a vector as difference of position vectors,so

angles $ bcz = cab,$ so

$$ bc \cdot cz / ( |bc|\, |cz| ) = ca \cdot ab / ( |ca|\, |ab| ) $$

  • $\begingroup$ "Angles $bcz=cab$", how did you get that, if its a property then which one is it. $\endgroup$
    – user350331
    Jul 23, 2016 at 19:47
  • $\begingroup$ indicated as above, angles in alternate segment of a circle equal. Vector bc = c-b etc.. $\endgroup$
    – Narasimham
    Jul 23, 2016 at 19:51
  • $\begingroup$ But how would the value of $|bc|, |cz|, |ca|$ and $|ab|$ be computed, moreover including $|cz|$ in the equation made the equation harder to solve. $\endgroup$
    – user350331
    Jul 23, 2016 at 21:48

Consider $O$ to be the origin, then $|a|=|b|=|c|=r$

We know that the equation of a line passing through points $z_1$ and $z_2$ is represented by

$$\begin{vmatrix} z & \overline{z} & 1 \\ z_1 & \overline{z_1} & 1 \\ z_2 & \overline{z_2} & 1 \\ \end{vmatrix}= 0$$

Now as per the question $a, b$ and $c$, lie on a circle, now lets consider a dummy point $d$ also located on the circumfrence of the circle. Then the equation of chord passing through $a$ and $b$ is represented as

$$\begin{vmatrix} z & \overline{z} & 1 \\ a & \overline{a} & 1 \\ b & \overline{b} & 1 \\ \end{vmatrix}= 0$$

which on simplification gives $$z(\overline{a}-\overline{b})-\overline{z}(a-b)+(\overline{a}b-a\overline{b})=0 \tag{1}$$

Similarly, equation of the chord passing through $c$ and $d$ can be given by

$$z(\overline{c}-\overline{d})-\overline{z}(c-d)+(\overline{c}d-c\overline{d})=0 \tag{2}$$

On eliminating $\overline{z}$ from equations $(1)$ and $(2)$, we get

$$z=\dfrac{(a-b)(\overline{c}d-c\overline{d})+(\overline{a}b-a\overline{b})(c-d)}{(\overline{a}-\overline{b})(c-d)+(\overline{c}-\overline{d})(a-b)}$$ Now, as $|a|^2=|b|^2=|c|^2=r^2$, so for every $\overline{z}, z\in\{a,b,c\}$, we can write $\overline{z}=\dfrac{r^2}{z}$

On simplification we get $$z=\dfrac{a^{-1}+b^{-1}-c^{-1}-d^{-1}}{a^{-1}b^{-1}-c^{-1}d^{-1}}$$

Now as the question stated that there is tangent at $c$ not a chord so we substitute $d=c$ to get the answer $$z=\dfrac{a^{-1}+b^{-1}-2c^{-1}}{a^{-1}b^{-1}-c^{-2}}$$


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