Annulus

$P$ is the midpoint between $Q$ and $S$ , so that its coordinates are $P\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ . Thus $||OP||^2=\dfrac{(x_{1} + x_{2})^2 + (y_{1} + y_{2})^2}{4}=r^2$ and the answer is $\boxed{4}$ .

Let $r$ and $R$ be the radius of the inner circle and larger circle respectively.

Let $P(r\cos(\theta),r\sin(\theta))$ and $x^2 + y^2 = r^2$

To find the tangent line at $P$ :

$\dfrac{dy}{dx}|_{P} = -\cot(\theta) \implies y = \dfrac{r - \cos(\theta)x}{\sin(\theta}.$

To find points $Q$ and $S$ we solve the system:

$x^2 + y^2 = R^2$

and,

$y = \dfrac{r - \cos(\theta)x}{\sin(\theta)}$

$\implies x^2 - 2r\cos(\theta)x + r^2 - R^2\sin^2(\theta) = 0 \implies$ $x = r\cos(\theta) \pm \sqrt{R^2 - r^2}\sin(\theta)$

Let $x_{1} = r\cos(\theta) + \sqrt{R^2 - r^2}\sin(\theta)$ and $x_{2} = r\cos(\theta) - \sqrt{R^2 - r^2}\sin(\theta)$

$\implies y_{1} = r\sin(\theta) - \sqrt{R^2 - r^2}\cos(\theta)$ and $y_{2} = r\sin(\theta) + \sqrt{R^2 - r^2}\cos(\theta)$

$\implies \dfrac{(x_{1} + x_{2})^2 + (y_{1} + y_{2})^2}{r^2} = \boxed{4}$ .