Consider a conic $X^2 + Y^2 - Z^2 = 0$ in $\mathbb{C}P^2$. In an affine chart $Z \neq 0$ it is supposed to look like a circle (however it looks in $\mathbb{C}^2$), but the deceptiveness of imagining it as something analogous to a real circle broke down when I realized that in this chart it has two points at infinity, namely $A = [1, i, 0]$ and $B = [1, -i, 0]$.
Moreover, it turns out that these points at least lie in the same analytic connected component as $C = [1, 0, 1]$, as can be seen by considering another affine chart $X \neq 0$, where by letting $y = Y/X$ and $z = Z/X$ we readily get $A = (i : 0)$, $B = (-i : 0)$, and $C = (0 : 1)$, so there is an obvious analytically smooth path $y = ki$, $z = \sqrt{1 - k^2}$ ($-1 \leq k \leq 1$) connecting the three points.
At this point cannot imagine what the complex flat conic looks like and how to visualize it. In the affine chart $Z \neq 0$ when $X/Z$ and $Y/Z$ are real it forms a familiar circle, but at the same time when, say, $Y/Z$ is purely imaginary, it now looks like a hyperbola, and when both are purely imaginary, no points satisfy the equation - and all this is just one chart! I assume that in other charts the picture is kind of similar: we get either real circles or real hyperbolas along different axes (and I assume that at some clever sections we get real parabolas), but visualizing it is hard.
UPD: the added complication is that my naive attempts at deforming a circle to a hyperbola fail: assuming $x = \hat x$, $y = e^{i\pi t/2} \hat y$, where $\hat x, \hat y \in \mathbb{R}$ the equation $x^2 + y^2 = 1$ becomes $\hat x^2 + e^{i \pi t}\hat y^2 = 1$, and this equation only has solutions when $t$ is half-integer.
My question is;
Is there a trick to understanding the conic better? Maybe something involving 3-spheres?