In this problem of a moving point in a graph, why is $aθ+θ/3=π/2$ being used to find the angle in which the points $O$,$P$, and $Q$ meet? In this problem the circles $C_1:x^2+y^2=1$ and $C_2:x^2+y^2=4$ are given along with the points $P(\cos(a\theta), \sin(a\theta))$ and $Q(2\cos(\frac{\pi}2-\frac{\theta}3), 2\sin(\frac{\pi}2-\frac{\theta}3)) $, which are on the circunference of the circles respectively.
The problem asks you to find the first value of $\theta$ in which the origin and the points $P$ and $Q$ will meet.
In the explanation, this image is given:
Representation of $C_1$ and $C_2$
and it is stated that when the blue and the red line meet $a\theta+\frac{\theta}3=\frac{\pi}2$ and therefore $\theta=\frac{3}{6a+2}\pi$.
My question is why is $a\theta+\frac{\theta}3=\frac{\pi}2$ being used to find the angle in which the points O,P, and Q meet?(What I mean by this is why is $\frac{\theta}3$ being used instead of $-\frac{\theta}3$ and why is the equation equal to $\frac{\pi}2$ and not $0$ since it is also near the two lines).
I would also like to know if the constant of the trigonometric function is the starting point and the variable is the direction (For example in  $Q(2\cos(\frac{\pi}2-\frac{\theta}3), 2\sin(\frac{\pi}2-\frac{\theta}3)) $) the line starts at $\frac{\pi}2$ and goes right because the variable $-\frac{\theta}3$ is negative right?).
Thank you.

Since the points P, Q, O are collinear, so the area of the triangle formed by them will be zero.
$\mathrm{P\equiv\left(cos(a\theta),sin(a\theta)\right)\,\,\,\,\&\,\,\,\,Q\equiv\left(2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right),2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\right)}$
So,
$\left|\begin{array}{ccc}\mathrm{cos(a\theta)}&\mathrm{sin(a\theta)}&\mathrm{1}\\\mathrm{2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}&\mathrm{2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}&\mathrm{1}\\\mathrm{0}&\mathrm{0}&\mathrm{1}\end{array}\right|=0$
Expanding only 3rd row,
$\implies\,1\cdot\left|\begin{array}{cc}\mathrm{cos(a\theta)}&\mathrm{sin(a\theta)}\\\mathrm{2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}&\mathrm{2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)}\end{array}\right|=0$
$\mathrm{\implies\,2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\,cos(a\theta)-2\,sin(a\theta)\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)=0}$
$\mathrm{\implies\,2\left\{sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\,cos(a\theta)-cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\,sin(a\theta)\right\}=0}$
$\mathrm{\implies\,2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}-a\theta\right)=0}$
$\mathrm{\implies\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}-a\theta\right)=0}$
Since the question asked to find the first value of $\,\,\theta\,\,$, so,
$\mathrm{\implies\,\dfrac{\pi}{2}-\dfrac{\theta}{3}-a\theta=0}$
$\mathrm{\implies\,\dfrac{\theta}{3}+a\theta=\dfrac{\pi}{2}}$
$\mathrm{\implies\,\left(\dfrac{1+3a}{3}\right)\theta=\dfrac{\pi}{2}}$
$\mathrm{\implies\,\theta=\dfrac{3}{1+3a}\cdot\dfrac{\pi}{2}}$
$\mathrm{\implies\,\theta=\dfrac{3\pi}{6a+2}}$
