A circle C _ 1 passes through the origin O and has diameter 4 on...

The centre of the circle ${C}_{1}$ will be $\left(2, 0\right)$ and radius is $2$

So, the equation of the circle ${C}_{1}$ is ${x}^{2} + {y}^{2} - 4x = 0$

Let θ be the angle made by the line $y = 2x$ with positive $x$-axis.

i.e. tanθ = 2

Now ${C}_{2}$ is a circle with $OA$ as diameter.

So, tangent at $A$ on ${C}_{2}$ is perpendicular to $OA$

Let $OA = l$

∴ \dfrac{QA}{AP} = \dfrac{lcotθ}{ltanθ}

\to \dfrac{QA}{AP} = \dfrac{1}{\tan^{2}θ} = \dfrac{1}{4}