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All vectors are in $\mathbb{R}^3$ and only $\mathbf{r} = \left[ x; y; z \right]$ is unknown. My question is does the following system define a conic section in the $x-y$ plane and, if so, how can I find it:

$$ \begin{align} \mathbf{r}^\mathrm{T} \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] & = 0 \\ \\ \mathbf{v}_1^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_1} {\Vert \mathbf{r} - \mathbf{r}_1 \Vert } + \mathbf{v}_2^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_2} {\Vert \mathbf{r} - \mathbf{r}_2 \Vert } & = c \end{align} $$

If either $\Vert \mathbf{v}_1 \Vert = 0$ or $\Vert \mathbf{v}_2 \Vert = 0$, then the above is the intersection of a cone and the $z=0$ plane. Likewise if $\mathbf{r} = \mathbf{r}_1$ or $\mathbf{r} = \mathbf{r}_2$. However, I have been unable to figure out what the above represents in the general case.

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