If you have a loop of string, a fixed point and a pencil, and stretch the string as much as possible, you draw a circle. With 2 fixed points you draw an ellipse. What do you draw with 3 fixed points?
Assume that three non-collinear points $A$, $B$, $C$ in the plane are given, and that you have a loop of string of length $\ell>|AB|+|BC|+|CA|$. Slinging this string around the three points and a pencil you can draw a loop $\gamma$ around $\triangle(ABC)$ in the obvious way, keeping the string tight at all times. This loop will be a continuous curve. In order to describe $\gamma$ more precisely we draw the lines $A\vee B$, $B\vee C$, $C\vee A$ in full. In this way the plane is divided into $7$ compartments, one of them the triangle $\triangle(ABC)$. The loop $\gamma$ traverses the $6$ unbounded compartments, and at each intersection with one of the above lines it has a corner. Within a given compartment $\gamma$ is an arc of an ellipse, whereby two of the three vertices $A$, $B$, $C$ act as foci. The points $A$, $B$ are foci for two such arcs. One of these arcs belongs to an ellipse with major axis $\ell-|AB|$ and the other to an ellipse with major axis $\ell-|BC|-|CA|$.
Let length of sting be > max(AB,BC,CA). If the given three points are A,B and C are not in a straight line, then
there are three possibilities to trace three ellipses and the locus is composed of the these three discontinuous ellipses.
Center point B can be ignored when inter-focal distance is AC,
Center point C can be ignored when inter-focal distance is BA, and
Center point A can be ignored when inter-focal distance is BC.
So in effect each is only a 2-point curve.