complex-geometrygeometrylinear algebra
In some intersection problems, like the 2D circle-circle intersection, there are two possible solutions that arise from a quadratic equation. If the circles do not intersect on the cartesian plane, the intersection points become complex numbers, because the discriminant of the quadratic is negative.
Can these complex intersection points be visualized spatially? And what is the significance of the magnitudes and phases of these complex numbers?
The answer given by Seyhmus is fine if the line is a line through the origin (or $d = 0$ in your formula). Perhaps a better solution, in the general case, is to find the interserction of a circle $|c - x|^2 = r^2$ with a parametric line, $$ x(t) = a + tb $$ where $a$ is a point and $b$ is a vector.For a point on this line to satisfy the equation, you need to have $$ (tb + (a-c)) \cdot (tb + (a-c)) = r^2 $$ which is a quadratic in $t$: $$ |b|^2 t^2 + 2(a-c)\cdot b t + (|a-c|^2 - r^2) = 0 $$ whose solutions are $$ t = \frac{-2(a-c)\cdot b \pm \sqrt{[2(a-c)\cdot b]^2 - 4|b|^2(|a-c|^2 - r^2)}}{2|b|^2} $$ Once you find these two values of $t$ and plug them back intot he parametric equation, you get your two points.
Coincidentally, this method works fine in all dimensions (i.e., $c$ and $a$ can be points of $R^3$, and $b$ a vector in $R^3$), and the same formula works.
In two dimensions, it's pretty easy to convert from the implicit form of the line to the parametric form: find any two points $P$ and $Q$ on the line, and let $a = P$ and $b = (Q - P)$. In three dimensions, it's rather difficult to specify a line implicitly, but quite simple to express it parametrically, so the general case tends to be asked in exactly the form in which I've answered it.
The operation that takes a point $P$ on the circle and find the other intersection point of the line $(AP)$ with the circle is a real automorphism of the circle.
The composition of $3$ such automorphisms is still a real automorphism. If you identify the circle with $\Bbb P^1(\Bbb R)$, it corresponds to a real homography $t \mapsto \frac {at+b}{ct+d}$.
In theory, given $3$ points on the circle and their images you can determine the parameters $a/d,b/d,c/d,\ldots \in \Bbb P^1(\Bbb R)$ and from then construct the line going through the two fixpoints (the parameters of the line are rational fractions in $a,b,c,d$ so they're a rational fraction in the coordinates of the three image points).
There probably even is a way to get a construction that doesn't depend on which $3$ points you choose on the circle, but at the moment I haven't checked on how to construct that line.
Given a point $A \in \Bbb P^2(\Bbb R)$, call $\sigma_A$ the involution of the circle obtained by taking a point $M$ and intersecting the line $(MA)$ with the circle.
Say an automorphism of the circle is direct if the output points turn in the same direction as the input point and indirect if it's not the case (this given by the sign of the determinant $ad-bc$).
The $\sigma_A$ are indirect involutions so they don't represent all the indirect automorphisms (a space of dimension $3$ while we have only $2$ dimensions by picking $A$). However, any direct automorphism can be given in infinitely many ways as a composition of two $\sigma_{A_i}$.
The fixpoints (possibly over $\Bbb C$) of $\sigma_A \circ \sigma_B$ are clearly the (possibly complex) intersections of $(AB)$ with the circle, so for any $C \in (AB)$ there is a unique point $D$ such that $\sigma_A \circ \sigma_B = \sigma_C \circ \sigma_D$. Constructing $D$ or $C$ from the other one is really easy, for example $C$ is the intersection of $(AB)$ with the line $(\sigma_D(M))( \sigma_A \circ \sigma_B (M))$, for any $M$ on the circle.
Next, we clearly have the simplification $\sigma_A \circ \sigma_A = id$.
So given $4$ points we can simplify $(\sigma_A \circ \sigma_B) \circ (\sigma_C \circ \sigma_D)$ by looking at the intersection of $(AB)$ and $(CD)$ (which is possibly at infinity) and moving both $B$ and $C$ to that point while changing $A$ and $D$ appropriately to $A'$ and $D'$. This gives us $(\sigma_A \circ \sigma_B) \circ (\sigma_C \circ \sigma_D) = \sigma_{A'} \circ \sigma_{D'}$
So now that we know how to represent direct automorphisms of the circle with couples of points and compute compositions, and because there is a direct relation between the fixpoints and the line formed by the points, all we have to do is to compute $\sigma_C \circ \sigma_B \circ \sigma_A \circ \sigma_C \circ \sigma_B \circ \sigma_A$ and simplify it to some $\sigma_X \circ \sigma_Y$. Then $(XY)$ intersects the circle at the two fixpoints $D_1$ and $D_2$ of $\sigma_C \circ \sigma_B \circ \sigma_A$.
Note that this construction should work if you replace the circle with any nondegenerate conic, so ellipses, parabolas, and hyperbolas too. You can also generalize this to an odd number $n$ of points instead of $3$, though the complexity of the construction increases linearly with $n$.
Best Answer
To be precise: the coordinates of the points of intersection become complex numbers. More precisely, they will form a conjugate pair of complex points, i.e. if $(a+ib,c+id)$ is one such solution, then you know the other solution to be $(a-ib,c-id)$ (if your circles were real).
Connecting these points you get the radical axis of the circle, just as you get for real points of intersection. You can imagine parametrizing that radical axis with a single complex parameter. To make this more precise, you could say that the line connecting the centers will intersect the radical axis at a point $t=0$, and from there a unit distance step along the radical axis will take you to a point $t=1$. You have to arbitrarily choose a direction for the axis here. You could come up with a formula to compute the point on the axis for every $t\in\mathbb C$.
Now you could draw a 2d diagram of that radical axis alone. You'd have the real and imaginary component of $t$ as two coordinates. A complex pair of points would be mirror images with respect to the real axis. Furthermore, for reasons of symmetry, the real component of your points of intersection would have to be zero in this coordinate system. In other words, the real coordinate of the solutions in the original coordinate system is the point where the radical axis meets the line joining the circle centers.
Visualizing two points on the imaginary axis all by themselves is not that interesting. Things do become more interesting if your setup is dynamic in some way. For example in this slide of a talk on a project I'm involved with, one can see the $x$ coordinates of the points you get by intersecting a circle with a horizontal line which you can move around. As you move the line from an intersecting position to a non-intersecting one, the points will come closer together along the real axis, and after meeting at the origin they will drift apart along the imaginary axis. (The slide in question also illustrates how to avoid the singular situation, so the points there won't move through the center but curve around it instead.) The picture of two circles intersecting would look pretty much the same in the parameter space of the radical axis I described above.