1. Use the geometry of the circle

Since arc $PQ$ subtends a right angle at the centre $O$, we have $\angle POQ = 90^\circ.$

The midpoint $R$ of the arc $PQ$ implies that $OR$ bisects the central angle $\angle POQ$. Hence, $\angle POR = \angle ROQ = 45^\circ.$

Also, all radii are equal, so $|\vec u|=|\vec v|=|\overrightarrow{OQ}|=r.$

2. Express $\vec v$ in terms of $\vec u$ and $\overrightarrow{OQ}$

Since $\vec v=\overrightarrow{OR}$ makes equal angle with $\overrightarrow{OP}$ and $\overrightarrow{OQ}$, it lies along the angle bisector of the two equal-length vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$.

Hence, $\vec v \parallel \vec u + \overrightarrow{OQ}.$

So, $\overrightarrow{OQ} = \lambda \vec v - \vec u$ for some scalar $\lambda$.

But given $\overrightarrow{OQ}=\alpha \vec u+\beta \vec v,$ comparing, we get $\alpha=-1,\qquad \beta=\lambda.$

So we only need to find $\beta$.

3. Find $\beta$ using magnitude

Since $|\overrightarrow{OQ}|=r$, we use $\overrightarrow{OQ}=\beta \vec v-\vec u.$

Now, $|\beta \vec v-\vec u|^2=r^2.$

Using $|\vec u|=|\vec v|=r$ and angle between $\vec u$ and $\vec v$ equal to $45^\circ$, $\vec u\cdot \vec v = r^2\cos 45^\circ = \frac{r^2}{\sqrt2}.$

Therefore, \begin{align*} |\beta \vec v-\vec u|^2 &= \beta^2|\vec v|^2+|\vec u|^2-2\beta(\vec u\cdot \vec v) \ &= \beta^2 r^2 + r^2 -2\beta\cdot \frac{r^2}{\sqrt2}. \end{align*}

Set this equal to $r^2$: $\beta^2 r^2 + r^2 - \sqrt2\beta r^2 = r^2.$

So, $\beta^2 - \sqrt2\beta = 0,$ which gives $\beta=\sqrt2$ (nonzero value, since $\overrightarrow{OQ}\neq -\vec u$).

Thus, $\beta^2 = 2.$

4. Identify the roots

We have found $\alpha=-1,\qquad \beta^2=2.$

So the required quadratic has roots $-1$ and $2$.

Hence, $x^2-(\text{sum of roots})x+(\text{product of roots})=0$ $x^2-(1)x+(-2)=0$ $x^2-x-2=0.$

5. Match with the options

This is Option D.

Derived answer: $\boxed{D}$