Locus of $\frac{|z-a|}{|z-b|}=c, c<1$ is a circle

Question

I saw similar questions with this one, but I did not find geometrical solution. Therefore, I want to share my solution and also seek other simpler solutions.

Problem:

I need to prove that the locus of $\frac{|z-a|}{|z-b|}=c, c<1$ is a circle.

My solution:

1. Define point $p$ such that $a,b,p$ are collinear and $\frac{|p-a|}{|p-b|}=c^{2}$
2. Define point $q$ between $a,b$ such that $\frac{|q-a|}{|q-b|}=c$
3. Define any point $z$ such that $|z-p|=|q-p|$

please refer to the picture

Upon checking, triangles $paz$ and $pbz$ are similar. $\frac{|p-a|}{|p-z|}=\frac{|p-z|}{|p-b|}=\frac{|z-a|}{|z-b|}=c$

Because of the way we defined the points above, the locus of $\frac{|z-a|}{|z-b|}=c$ is the circle $|z-p|=|p-q|$

What do You think? Please post Your solutions too, thanks!

Answer 1

Here is the proof I learned back in the day (in geometrical notation, not complex arithmetic, because I'm more used to that in this context):

Let $r$ be a point on the line $ab$ to the left of $a$ such that $\frac{|ra|}{|rb|} = c$. Since $\frac{|zb|}{|za|}=\frac{|qb|}{|qa|}$, the line $zq$ is the angle bisector of $\angle azb$. Similarly, we get that the line $zr$ is the angle bisector of the external angle at $z$ (i.e. is the bisector of the supplement of $\angle azb$).

(It is a known fact that angle bisectors in a triangle are the lines from one corner to the opposite side that divides the opposite side into pieces that have the same ratio as the adjacent sides. This is also true for the external bisectors.)

This makes $\angle qzr=90^\circ$, meaning $z$ lies on the circle with diameter $qr$.

Answer 2

An algebraic solution: square everything, write $z=x+yi$, use $|w|^2 = Re(w)^2 + Im(w)^2$, rearrange everything, and voila! The equation for a circle should pop out.

Greg