Hint: It is a ellipse. To see this suppose that you have $A=(x,0)$ and $B=(0,y)$, so:
$$
x^2+y^2=4
$$
Also there are tow points that divide the segment $AB$ internally by $1:2$ and these are $(s,t)=(\frac{x}{3},\frac{2y}{3})$ and $(z,w)=(\frac{2x}{3},\frac{y}{3})$. Now you can replace $x,y$ in the previous equation by $s,t,z,w$:
$$
x^2+y^2=9s^2+\frac{9t^2}{4}=4
$$
and
$$
x^2+y^2=\frac{9z^2}{4}+9w^2=4
$$
which are two ellipses.
I think you overworked here. The condition is
$$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$
$$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$
and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3a}2\;$ , and since radius = absolute value of $\;x$ - coordinate of circle, the circle is tangent to the $\;y$ - axis
Let $\;Q=(x,y)\;$ , then its distance from the $\;y$ - axis is $\;|x|\;$ , and the length of any of its two tangets to the above circle is
$$\sqrt{\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4}$$
and thus we get the equation:
$$x^2=\left(x-\frac{3a}2\right)^2+y^2\;-\;\frac{9a^2}4\iff$$
$$\iff -3ax+y^2=0\;\;(*)$$
Now, let us check what is the locus of all point whose distance to $\;4x+3a=0\iff x=-\frac{3a}4\;$ equals its distance to $\;\left(\frac{3a}4\,,\,0\right)\;$:
$$\left(x-\frac{3a}4\right)^2+y^2=\frac{(4x+3a)^2}{16}\iff x^2-\frac{3a}2x+\frac{9a^2}{16}+y^2=x^2+\frac{3a}2x+\frac{9a^2}{16}\iff$$
$$\iff -3ax+y^2=0\;\;(*)$$
and thus both conditions are identical
Best Answer
There is a rather elegant solution to this, but requires some construction, and would be difficult to include in an answer, but I shall try my best.
Now, in the above diagram, $P$ is the moving point, and $C$ replaces $A$ in your question (sorry about that, software is new to me). When you deflect $P$ w.r.t to $CP'$ by $\theta$, the length of $CP = 2r\cos\theta$
Now, we know that $BQ = k.CP$ from the condition of our problem, hence
$$BQ = 2kr \cos \theta$$ $$BQ' = 2kr$$
Hence, we will always have
$$BQ = BQ' \cos \theta$$
And since $Q'$ is a fixed point for a given $C$ and $B$, this would be a circle. Actually a pair, because it could be on either side of $BP$. I hope this answers your question