 Solution

No finite choice of $b$ and $c$ gives a line. The coefficient of $z{z}^{\ast }=x^2+y^2$ is fixed at one, so completing the square always gives a circle, a point, or the empty set. A general equation $Az{z}^{\ast }+b{z}^{\ast }+b^{\ast }z+c=0$ could describe a line when $A=0$ and $b\ne 0$, but changing the fixed quadratic coefficient is not permitted here. A limiting sequence of increasingly large circles does not make any one of these finite equations a line.