Proving $\sum_{k=1}^{4} b_k |z_k|^2=0$ if $z_k$ are con-cyclic s. t. $\sum_{k=1}^4 b_k =0=\sum_{k=1}^{4} b_k z_k$ If four complex numbers  $z_k$ are con-cyclic in Argand plane such that $\sum_{k=1}^4 b_k =0=\sum_{k=1}^{4} b_k z_k$ and $b_k$ are real, earlier it has been proved that
$$b_1b_2|z_1-z_2|^2=b_3 b_4|z_3-z_4|^2~~~~~(1)$$
after a good number of steps.
See Condition for cyclic quadrilateral given $\sum_{i=1}^{4}{b_i}=0$ and $\sum_{i=1}^{4}{b_iz_i}=0$
In a rather simple way one can prove that
$$\sum_{k=1}^{4} b_k |z_k|^2=0,~~~~(2)$$
Since $z_k$ are con-cyclic $|z_k-z_0|^2=R^2$, where let $z_0$ denote the center of the circle. Then we can write $$0=\sum_{k=1}^{4} b_k=\sum_{k=1}^{4} b_k R^2=\sum_{k=1}^4 b_k|z_k-z_0|^2 \implies \sum_{k=1}^{4} b_k [|z_k|^2+|z_0|^2-2 \Re (\bar z_0 z_k)]=0$$ $$ \implies \sum_{k=1}^{4} b_k |z_k|^2-\sum_{k=1}^42 b_k \Re(\bar z_0  z_k)=0$$
as $\sum_{k=1}^{4} b_k=0$
$$\implies \sum_{k=1}^4 b_k |z_k|^2-2\Re \left(\bar z_0 \sum_{k=1}^{4} b_k z_k\right)=0  \implies \sum_{k=1}^4 b_k |z_k|^2=0.$$
The question is how else the result (2) can be proved using (1) and its proof or otherwise.
 A: The following will prove that $(1) \Leftrightarrow (2)$ under the given assumptions.
To simplify the calculations, first note that $(2)$ is invariant to translations  $z \to z-w$ since $\sum b_k|z_k - w|^2=\sum b_k|z_k|^2$ following an argument similar to the one used in the original post. The assumptions are obviously invariant to translations, and so is $(1)$.
Also, both assumptions and $(1),(2)$ are invariant to complex multiplication $z \to z \cdot r\,e^{i \varphi}$.
It can then be assumed WLOG that $z_4 = 0$ and $z_3=1$ by combining a translation that makes $z_4$ the origin, and a rotation with appropriate scaling that brings $z_3$ to the unit point on the real axis.
With that, the assumptions reduce to:
$$
b_1 + b_2 + b_3 = -b_4 \tag{a}
$$
$$b_1 z_1 + b_2 z_2 = -b_3 \tag{b}
$$
And the relations $(1),(2)$ reduce to:
$$
b_1b_2|z_1-z_2|^2 = -b_3(b_1+b_2+b_3) \tag{1}
$$
$$
b_1|z_1|^2+b_2|z_2|^2 + b_3 = 0 \tag{2}
$$
Multiplying assumption $(b)$ by its conjugate gives:
$$
\begin{align}
b_3^2 &= \left(b_1z_1+b_2z_2\right)\left(b_1\bar z_1+b_2\bar z_2\right)
\\ &= b_1^2|z_1|^2 + b_2^2|z_2|^2 + 2 b_1b_2 \Re{(z_1\bar z_2)}
\end{align}
$$
$$
\implies 2 b_1b_2 \Re{(z_1\bar z_2)} = b_3^2 - b_1^2|z_1|^2 - b_2^2|z_2|^2
$$
Then:
$$
\begin{align}
b_1b_2|z_1-z_2|^2 &= b_1b_2\left(|z_1|^2+|z_2|^2-2\Re{(z_1\bar z_2)}\right)
\\ &= b_1b_2|z_1|^2 + b_1b_2|z_2|^2 - \left(b_3^2 - b_1^2|z_1|^2 - b_2^2|z_2|^2\right)
\\ &= \left(b_1b_2|z_1|^2+b_2^2|z_2|^2\right) + \left(b_1^2|z_1|^2+b_1b_2|z_2|^2\right) - b_3^2
\\ &= b_2\left(b_1|z_1|^2+b_2|z_2|^2\color{red}{+b_3-b_3}\right)+b_1\left(b_1|z_1|^2+b_2|z_2|^2\color{red}{+b_3-b_3}\right) - b_3^2
\\ &= (b_1+b_2)\left(b_1|z_1|^2+b_2|z_2|^2+b_3\right) - b_3\left(b_1+b_2+b_3\right)
\end{align}
$$
It follows that $(1) \Leftrightarrow (2)$ when $b_1+b_2 \ne 0$, and the case $b_1+b_2 = 0$ can be easily verified.
