Shape induced by the condition of angle bisector passing through a fixed point

Question

Let $A$, $B$ be two points on 2D plane. For any $C \in \overline{AB}$, define the set $$ S = \{P \; | \; \angle APC = \angle CPB \}.$$ What is the shape of $S$?

I was thinking about graphics defined with simple conditions involving two fixed points on 2D plane for example sum of $\overline{AP}+\overline{PB}$ is a constant gives you a eclipse, $\angle APB$ is a constant gives you two parts of circles combined together, $\overline{AP}-\overline{PB}$ is a constant gives you a hyperbola.

For this particular question it is equivalent to ask(by the angle bisector theorem) the points $P$ such that $\overline{AP} : \overline{PB} = \overline{AC} : \overline{CB}$, where $ \overline{AC} : \overline{CB}$ is a constant. I have no further idea.

Answer 1

Hints: We may use this property that in triangle PAB perpendicular bisector of AB meet the bisector of $\angle APB$ on the circumcircle of ABC at a point like D. So The locus of P can be a circle as can be seen in figure. All lines PD cross AB at points like C such that :

$$\angle APC=\angle CPB$$

May be using analytic geometry we can find the equation locus.

Answer 2

Here is a solution using coordinate geometry. If $|AB| = k$ and $\dfrac {|AC|}{|BC|} = t$ where $C \in AB$.

Using angle bisector theorem,

$\displaystyle \frac{|AP|}{|PB|} = \frac{|AC|}{|BC|} = t \implies |AP|^2 = t^2 \cdot |PB|^2$

WLOG, coordinates of $A$ is $(0, 0)$ and of $B$ is $(k, 0)$. If coordinates of $P$ is $(x, y)$,

we get $~x^2 + y^2 = t^2 ((x-k)^2 + y^2)$

If $ \displaystyle t = 1, x = \frac{k}{2}$ i.e. the locus is perpendicular bisector of $AB$.

If $t \ne 1,~ \displaystyle x^2 + y^2 - \frac{2 k t^2}{t^2-1} x = - \frac{k^2 t^2}{t^2-1}$

$ \displaystyle \left(x - \frac{kt^2}{t^2-1}\right)^2 + y^2 = \left(\frac{kt}{t^2-1}\right)^2$

which is a circle with center at $ \displaystyle \left(\frac{kt^2}{t^2-1}, 0 \right)$ and radius $ \displaystyle \frac{kt}{|t^2-1|}$