Zero potential locus of two charges in 2D plane

Suppose we place two charges $$q_{1}$$ and $$q_{2}$$ in the cartesian plane in the points $$A(x_{1},0)$$ and $$B(x_{2},0)$$ and we want to find the locus where the potential $$V$$ of the field is zero (Notice that in order to simplify the problem we put the two charges lying in the x axis).

Without any loss of generality we can assume that $$q_{1} = \lambda q_{2}$$ (1), where λ is a costant. In order for the net potential of the field to give zero, the two charges need to be of opposite signs and thus we conclude that λ in equation (1) is a negative constant. Consequently, it is true that $$\lambda\in(-\infin,0)$$(2).

Now, lets assume there is a point $$M(x,y)$$ in 2D space which satisfies the condition that its potential is zero. From the superposition principle it is true that:$$V_{(M)}=V_{(A)\rightarrow(M)}+V_{(B)\rightarrow(M)}=0$$which expands to:$$V_{(M)}=K\frac{q_{1}}{(AM)}+K\frac{q_{2}}{(BM)}=0$$.From the assumption (1) that we introduced, the equation above transforms to:$$V_{(M)}=K(\frac{q_{1}}{(AM)}+\frac{q_{2}}{(BM)})=Kq_{1}(\frac{1}{(AM)}-\frac{|\lambda|}{(BM)})=0\qquad(3)$$$$K$$ and $$q_{1}$$ are non-zero constants so equation (3) becomes:$$(BM)=|\lambda|(AM) \ \ (4)$$It is true that:

$$(AM)=\sqrt{(x-x_{1})^2+y^2}\ and \ (BM)=\sqrt{(x-x_{2})^2+y^2}$$and (4) transforms to:$$\sqrt{(x-x_{2})^2+y^2}=|\lambda|\sqrt{(x-x_{1})^2+y^2}$$Now both sides of the equation above are positive and thus we can square both sides:$$(x-x_{2})^2+y^2=\lambda^2[(x-x_{1})^2+y^2]\Leftrightarrow(1-\lambda^2)x^2+(1-\lambda^2)y^2+2x(\lambda^2x_{1}-x_{2})+x_{2}^2-\lambda^2x_{1}^2=0 \quad (5)$$If $$\lambda\in(-\infin,-1)\cup(-1,0)$$ (5) becomes:$$x^2+y^2+2(\frac{x_{2}-\lambda^2x_{1}}{1-\lambda^2})x+(\frac{x_{2}^2-\lambda^2x_{1}^2}{1-\lambda^2})=0 \quad (6)$$Equation (6) describes a circle (in the form $$x^2+y^2+Ax+C=0$$) with its centre moving in the x' axis.

Its centre K is$$K\equiv(-A/2, 0)\equiv(\frac{\lambda^2x_{1}-x_{2}}{\lambda^2-1},0)$$and its radius ($$r = \frac{\sqrt{A^2-4C}}{2}$$):$$r=|\frac{x_{2}-x_{1}}{\lambda+\lambda^{-1}}|=\frac{d(A,B)}{|\lambda+\lambda^{-1}|}$$Now we need to check the case of $$\lambda = -1$$:

From the equation (5) we get:$$2x(x_{2}-x_{1})+x_{2}^2-x_{1}^2=0\Leftrightarrow2x(x_{1}-x_{2})+(x_{2}-x_{1})(x_{1}+x_{2})=0\Leftrightarrow x=\frac{x_{1}+x_{2}}{2}$$So the zero potential locus is either a circle when $$\lambda\neq-1$$ or the perpendicular bisector of (AB) when $$\lambda=-1$$ and thus $$(q_{1}=-q_{2})$$.