Let $S = \{z \in \mathbb{C} : \left|\frac{z-6i}{z-2i}\right| = 1 \text{ and } \left|\frac{z-8+2i}{z+2i}\right| = \frac{3}{5}\}$. Then $\sum_{z \in S} |z|^2$ is equal to

The locus of the point representing the complex number $z$ for which $|z+3|^2-|z-3|^2=15$ is

If $z_{1}$ and $z_{2}$ are two non-zero complex numbers such that $\frac{z_{1}}{z_{2}}+\frac{z_{2}}{z_{1}}=1$, then the origin and the points represented by $z_{1}$ and $z_{2}$:

If ${Z_1} \ne 0$ and ${Z_2}$ are two complex numbers such that $\frac{{{Z_2}}}{{{Z_1}}}$ is a purely imaginary number,then $\left| {\frac{{2{Z_1} + 3{Z_2}}}{{2{Z_1} - 3{Z_2}}}} \right|$ is equal to

$P$ is a point denoting $z$ in the Argand diagram. If $\frac{z-i}{z-1}$ is always purely imaginary,then the locus of $P$ is

The equation $\text{Im}\left( \frac{iz - 2}{z - i} \right) + 1 = 0$,where $z \in \mathbb{C}$ and $z \neq i$,represents a part of a circle having radius equal to