If z₁, z₂, z₃ are non-zero, non-collinear complex numbers such that $\frac{2}{{\mathrm{z}}_1}=\frac{1}{{\mathrm{z}}_2}+\frac{1}{{\mathrm{z}}_3}$, then the points z₁, z₂, z₃ lie

We have,

$\begin{array}{l}\frac{2}{{\mathrm{z}}_1}=\frac{1}{{\mathrm{z}}_2}+\frac{1}{{\mathrm{z}}_3}=\frac{{\mathrm{z}}_3+{\mathrm{z}}_2}{{\mathrm{z}}_2{\mathrm{z}}_3}\\ \Rightarrow {\mathrm{z}}_1=\frac{2{\mathrm{z}}_2{\mathrm{z}}_3}{{\mathrm{z}}_2+{\mathrm{z}}_3}.\end{array}$

Now, $\left(\frac{{\mathrm{z}}_2-{\mathrm{z}}_4}{{\mathrm{z}}_1-{\mathrm{z}}_4}\right)\left(\frac{{\mathrm{z}}_1-{\mathrm{z}}_3}{{\mathrm{z}}_2-{\mathrm{z}}_3}\right)$

$=\left(\frac{z_2-z_4}{\frac{2z_2{z}_3}{z_2+z_3}-z_4}\right)\left(\frac{\frac{2z_2{z}_3}{z_2+z_3}-z_3}{z_2-z_3}\right)$

$=\frac{{\mathrm{z}}_2}{\frac{2{\mathrm{z}}_2{\mathrm{z}}_3}{{\mathrm{z}}_2+{\mathrm{z}}_3}}\frac{{\mathrm{z}}_3\left({\mathrm{z}}_2-{\mathrm{z}}_3\right)}{\left({\mathrm{z}}_2+{\mathrm{z}}_3\right)\left({\mathrm{z}}_2-{\mathrm{z}}_3\right)}$               [taking z₄ = 0]

$=\frac{1}{2}$ (a real number).

Hence, points z₁, z₂, z₃ and origin are concyclic and therefore, z₁, z₂, z₃ lie on a circle passing through the origin.