I have checked the existing question Intersection of two circles. and model for intersection of two circles in the complex projective plane - I do not think either of these answers my question.
The intersection of two circles (in the most generic case) produces four points. Two real points and two other points (1, +-i). These two other points are usually skipped since they have imaginary components. But it is critical in understanding how projective geometry compares to Euclidean geometry.
My question is, is there any way to visualize these two points? Usually one compares complex plane with a 2D Euclidean space, but this is not the same.
Can you help me interpret it figuratively? I will also use this intuition/visualization to help me understand the two circular points.
It depends on the background. If you are defining Euclidean geometry based on the imaginary circle points $(1,\pm i,0)$ (along the lines of a Cayley-Klein metric), then it is possible to replace these two complex points by two real points, e.g. $(1,\pm 1,0)$, and study the resulting pseudo-Euclidean geometry. I know Perspectives on Projective Geometry has some illustrations and explanations for this. In this world, “circles” would be hyperbolas with asymptotes of slope $\pm1$. Angles would not circle round after a full turn, but add up indefinitely just as lengths do (“hyperbolic measurement”). It's a strange world, but many things from Euclidean geometry have their counterpart in pseudo-Euclidean geometry.