Prove: minimal origin-to-ellipse $|z+a|+|z-a|=2r$ lies on minor axis, using complex numbers

Question

Let $|z+a|+|z-a|=2r$ where $a,z\in\mathbb{C}$ and $r>|a|$.

Prove the minimum of $|z|$ is $\sqrt{r^2-|a|^2}$

The equation describes an ellipse, and it's clear from drawing a picture that the minimal $|z|$ are the two vertices of the ellipse on the minor axis. How can we prove this claim analytically using complex numbers?

Thank you in advance!

Answer 1

Turns out the example I made up is correct: all the important points have integer real and imaginary parts. Draw $a = 20 + 15i$ and $r=65$

Answer 2

$|z+a|+|z-a|=2r$ is equation of an ellipse with foci at $-a$ and $a.$
Focal distance is $|2a|$ and the length of one axis is $2r.$ Thus the other axis has length $$2|d|=2\sqrt{r^2-|a|^2}.$$ Since the axes of ellipse are perpendicular, extremal points (vertices) lying on this axis are $$\pm d=\pm i\times |d|\times {a\over |a|}.$$
The nearest points to the origin are vertices of the ellipse that lie on the minor axis.

• If $|a|<\sqrt{r^2-|a|^2},$ then $z_{1,2}=\pm a.$
• If $|a|>\sqrt{r^2-|a|^2},$ then $z_{1,2}=\pm d.$