Question

If $b_1+b_2+b_3+b_4=0$ where $b_i\in R$ such that the sum of no two values being zero and $b_1{z}_1+b_2{z}_2+b_3{z}_3+b_4{z}_4=0$ where $z_1,z_2,z_3,z_4$ are arbitary complex numbers such that no three of them are collinear, so the required condition for complex numbers to be concylic is

• A. $|b_1{b}_4||z_1-z_4{|}^2=|b_2{b}_3||z_2-z_3{|}^2$
• B. $|b_1{b}_3||z_1-z_3{|}^2=|b_2{b}_4||z_2-z_4{|}^2$
• C. $|b_1{b}_2||z_1-z_2{|}^2=|b_3{b}_4||z_3-z_4{|}^2$
• D. None of these

Answer 1

The correct option is C $|b_1{b}_2||z_1-z_2{|}^2=|b_3{b}_4||z_3-z_4{|}^2$


$b_1+b_2=-(b_3+b_4),b_1{z}_1+b_2{z}_2=-(b_3{z}_3+b_4{z}_4)$
$\Rightarrow \frac{b_1{z}_1+b_2{z}_2}{b_1+b_2}=\frac{b_3{z}_3+b_4{z}_4}{b_3+b_4}$
Hence, the line joining the complex numbers $A\left(z_1\right),B\left(z_2\right)$ and $C\left(z_3\right),D\left(z_4\right)$ meet.
Let $P\left(z\right)$ be the point of intersection. These points will be concyclic, if $PA\times PB=PC\times PD$. Now,
$PA=\left|\frac{b_2}{b_1+b_2}\right||z_1-z_2|,PB=\left|\frac{b_1}{b_1+b_2}\right||z_1-z_2|$
$PC=\left|\frac{b_4}{b_3+b_4}\right||z_3-z_4|,PD=\left|\frac{b_3}{b_3+b_4}\right||z_3-z_4|$
$\Rightarrow |b_1{b}_2||z_1-z_2{|}^2=|b_3{b}_4||z_3-z_4{|}^2(\because |b_1+b_2|=|b_3+b_4|)$