If $\vert a-b \vert \vert c-w \vert = \vert a-c \vert \vert b-w \vert$ find the value of $w$ where $a,b,c,w$ are points in the complex plane

Question

Given the following equation $$\vert a-b \vert \vert c-w \vert = \vert a-c \vert \vert b-w \vert$$ find the value of $w$ where $a,b,c,w$ are points in the complex plane.

As per the equation seems, $w$ is the vertex of the triangle $\Delta CAW$ and $B$ is the point where the angle bisector intersects $AW$.

I couldn't really find out $w$ though I understand that it's a fixed point given the following configuration (I stated). This is how I tried : Map the points $c,a,b \mapsto 0,\frac{b-c}{a-c}, \frac {w-c}{a-c}$. Now, $\arg(\frac{b-c}{a-c}) = 2 \arg (\frac{b-c}{a-c})^2$. And then I know that the points $1, \frac{b-c}{a-c}, \frac{w-c}{a-c}$ are collinear. But anyway, I can't do anything further. Cannot really work out the value of $w$.

$w=a$ is a possible solution, but also the symmetric of $a$ wrt $bc$ is also a solution.. — Exodd, Jan 02 '18 at 13:29
Actually, there's an entire circle of solutions, called the Apollonius Circle https://en.wikipedia.org/wiki/Circles_of_Apollonius — Exodd, Jan 02 '18 at 13:31
@Mathbg See for example Proving the Apollonian circle formula. — dxiv, Jan 02 '18 at 16:25

score 1 · Answer 1 · edited Aug 19 '19 at 01:53

Call $AB,BW,WC,CA$ the segments on the complex plane that connect the points $a,b,w,c$. We know that $|a-b| = |AB|$, $|b-w| = |BW|$, etc.

The condition is

$$|AB|\cdot |CW| = |AC|\cdot|BW|$$ but the quantities $|AB|$ and $|AC|$ are fixed, so we are looking for points $w$ in the plane so that the distance from $c$ is $|AC|/|AB|$ times the distance from $b$.

The locus of such points is called Apollonius Circle. So there are infinite solutions to your problem, namely, every point on that circle.

If $\vert a-b \vert \vert c-w \vert = \vert a-c \vert \vert b-w \vert$ find the value of $w$ where $a,b,c,w$ are points in the complex plane

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