Interesting Locus problem

Question

A variable line passes through $P(2,-1)$ and cuts the co-ordinate axes at $A$ and $B$ respectively.

$Q$ lies on line AB such that $$\frac{2}{PQ} = \frac{1}{PA} + \frac{1}{PB}$$ Find the locus of point Q

Answer 1

$Q$ lies on a line through the origin with slope $\frac{1}{2}$.

Let we prove it through the properties of the circular inversion. We take a circle with centre at $P$ and radius $1$, then consider the inverse of the $x$-axis and the inverse of the $y$-axis. They are two circles ($\Gamma_x,\Gamma_y$) meeting at $P$. We have $A^{-1}\in\Gamma_x$ and $B^{-1}\in\Gamma_y$: in order to get $Q$, we just need the inverse of the midpoint of $A^{-1}B^{-1}$. But the midpoint of $A^{-1}B^{-1}$ always lies on a particular circle (the red circle in the above picture) of the coaxal bundle given by $\Gamma_x$ and $\Gamma_y$, i.e. the circle through $P$, the centre of $\Gamma_x$ and the centre of $\Gamma_y$. It follows that $Q$ always lies on a particular line, the line through $(2,1)$ and $(-2,-1)$.

That is also a consequence of the fact that the initial condition grants that the cross-ratio $(P,A,Q,B)$ is always the same. A third alternative is provided by the following fact: in a trapezoid having bases with lengths $a,b$, the segment parallel to the bases through the point of intersection of the diagonals has length given by the harmonic mean of $a$ and $b$, i.e. $\frac{2ab}{a+b}$.

Answer 2

I have taken a related case where P is the origin ( without loss of generality, the entire graph can be shifted to required point) with harmonic mean of two free segments (shown as dots $A,B$ , intersection points with radius vector) as constant (represented by the unit purple circle shown).