= Solution
For a real line $ux+vy+w=0$, with $(u,v)\ne(0,0)$, choose $B=(u-iv)/2$ and $C=w$. Direct multiplication gives $Bz+\overline B\,\overline z=ux+vy$, so \b[every line has the required form] with $B\ne0$.
For a <circle> of centre $c$ and radius $\rho>0$, expand its <modulus> equation:
$$
|z-c|^2=\rho^2
\iff z\overline z-\overline c z-c\overline z+|c|^2-\rho^2=0.
$$
Thus \b[the required <circle> parameters are] $\boxed{B=-c,\ C=|c|^2-\rho^2}$. Conversely, completing the square in the <circle> form gives $|z+B|^2=|B|^2-C$, which describes a genuine <circle> when $|B|^2-C>0$. The <complex conjugate> terms make both equations real-valued.
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