According to Single Axis Geometry by Fitting Conics by Jiang et al., one can compute the image of the circular points in a picture from conics which are the images of circles. Fit two conics to tracked points, then compute the intersection points of the two conics. There are always 4 intersection points including complex and infinite points according to Bezout’s theorem, of which 0-4 are real points. If there is a unique pair of complex conjugates, the image of circular points $\mathbf i$ and $\mathbf j$ are exactly this pair of complex conjugates. When there is no real intersection points we obtain two pairs of complex conjugate points and two possible solutions. The paper says this double solution ambiguity can be removed by using any additional conic.
My question is how can we do that?
Let's say we were able to obtain two conics $C_1$, $C_2$. As you can see here these conics have no real intersection points therefore we have two pairs of complex conjugate points:
\begin{align*} x_1 &= 1265.62 - 136.414i & y_1 &= 790.132 + 38.3403i \\ x_2 &= 1265.62 + 136.414i & y_2 &= 790.132 - 38.3403i \\ x_3 &= 334.378 + 136.414i & y_3 &= 790.132 + 38.3403i \\ x_4 &= 334.378 - 136.414i & y_4 &= 790.132 - 38.3403i \end{align*}
The paper says we can use any additional conic to resolve this problem. Let's add another conic $C_3$. The intersection points of $C_1$ and $C_3$ are:
\begin{align*} x_1 &= 800 - 9360.25i & y_1 &= 1696.11 - 0.00008i \\ x_2 &= 800 + 9360.25i & y_2 &= 1696.11 + 0.00008i \\ x_3 &= 800 - 2347.41i & y_3 &= 541.402 - 0.00002i \\ x_4 &= 800 + 2347.41i & y_4 &= 541.402 + 0.00002i \end{align*}
The intersection points of $C_2$ and $C_3$ are:
\begin{align*} x_1 &= 800 - 1767.11i & y_1 &= 446.355 - 0.0001i \\ x_2 &= 800 + 1767.11i & y_2 &= 446.355 + 0.0001i \\ x_3 &= 800.238 - 555946i & y_3 &= 91806.1 + 0.039i \\ x_4 &= 800.238 + 555946i & y_4 &= 91806.1 - 0.039i \end{align*}
Here is the picture of all three conics:
So how can we say what intersection points are the image of circular points?