 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 every line has the required form with $B\ne 0$.

For a circle of centre $c$ and radius $\rho >0$, expand its modulus equation:Thus 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.