$\text{ From a point }P\text{ representing complex number }{z}_1\text{ on the curve }\left(\right|z|=2)\text{ two tangents are drawn form }$

$P\text{ to the curve }\left|\mathrm{z}\right|=1\text{ meets at }\mathrm{A}\left({\mathrm{z}}_2\right)\text{ and }\mathrm{B}\left({\mathrm{z}}_3\right)\text{ then }$

$\begin{array}{l}OA=OB=1;OP=2\\ \sin \mathrm{\angle }OPA=\frac{1}{2}\mathrm{\angle }OPA=\frac{\pi }{6}\mathrm{\angle }APB=\frac{\pi }{3}\end{array}$

$\mathrm{OM}\perp \mathrm{AB}\text{ so }\mathrm{△}\mathrm{APB}\text{ is equilateral. so }\left(\mathrm{D}\right)\text{ is true. }$

$\mathrm{OM}=1/2;\mathrm{MN}=1/2\mathrm{PN}=1\text{ so }$

$\text{ centroid lies on }\left|z\right|=1----\left(A\right)$

$\begin{array}{l}\left|\frac{z_1+z_2+z_3}{3}\right|=1\\ {\left|z_1+z_2+z_3\right|}^2=9\\ \left(z_1+z_2+z_3\right)\left(\overline{z_1}+\overline{z_2}+\overline{z_3}\right)=9\\ Since z_1\overline{z_1}=4, z_2\overline{z_2}=1 and z_3\overline{z_3}=1\\ \text{ ⇒ }\left(\frac{4}{\overline{z_1}}+\frac{1}{\overline{z_2}}+\frac{1}{\overline{z_3}}\right)\left(\frac{4}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}\right)=9----\left(B\right)\\ \mathrm{\angle }\mathrm{AOB}=\pi -\frac{\pi }{3}=\frac{2\pi }{3}---\left(C\right)\end{array}$