General Solution of $|z-a| + |z-b| = c$

Question

If I plot $|z-a| + |z-b| = c$ on complex plane (where $z,a,b,c$ are complex numbers), what kinds of graphs should I expect and what is the general form of the solution?

I know that a circle can be obtained by say $|z+i|+|z-1|=2$ and an ellipse can be obtained by $|z - 3| + |z +3| = 10$. I am guessing that the general solution is conics sections.

Answer 1

If you want any solutions then $c$ must be a nonnegative real number, in which case you get a line segment from $a$ to $b$ if $c=|a-b|$ and an ellipse if $c>|a-b|$, and nothing otherwise. The circle you found is not actually a circle but an ellipse with foci at $i$ and $1$ that looks pretty circlelike. The only way to get a circle is to set $a=b$.

Answer 2

First of all, $c$ must be real and non negative, since it is the sum of two absolute values.

Second: an absolute value $|z-a|$ represents the distance between $a$ and $z$.

So the equation is the locus of points $z$ whose sum of distances from $a$ and $b$ is constantly equal to $c$, i.e., if not empty, an ellipse of foci $a$ and $b$.