An arc $P Q$ of a circle subtends a right angle at its centre $O$. The mid-point of an $\operatorname{arc} P Q$ is $R$. So, $P R=R Q$

Also given that, $$\overrightarrow {OP} = \overrightarrow u ,\overrightarrow {OR} = \overrightarrow v $$, and $$\overrightarrow {OQ} = \alpha \overrightarrow u + \beta \overrightarrow v $$



Let $$\overrightarrow {OP} = \overrightarrow u$$

$$ \begin{aligned} & \overrightarrow{OQ}=\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v} \\\\ & \overrightarrow{O R}=\overrightarrow{v} \end{aligned} $$

Clearly, $\overrightarrow{q}=\hat{\mathbf{j}}$

$$ \overrightarrow{u}=\hat{\mathbf{i}}, \overrightarrow{v}=\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}} $$

Now, given, $\overrightarrow{q}=\alpha \overrightarrow{u}+\beta \overrightarrow{v}$

$$ \begin{array}{ll} \Rightarrow & \hat{\mathbf{j}}=\alpha(\hat{\mathbf{i}})+\beta\left(\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}}{\sqrt{2}}\right) \\\\ \Rightarrow & \hat{\mathbf{j}}=\mathbf{i}\left(\alpha+\frac{\beta}{\sqrt{2}}\right)+\frac{\beta}{\sqrt{2}} \hat{\mathbf{j}} \end{array} $$

On comparing both sides, we get

$$ \begin{array}{rlrl} \alpha+\frac{\beta}{\sqrt{2}} =0 \text { and } \frac{\beta}{\sqrt{2}}=1 \\\\ \Rightarrow \alpha =-\frac{\beta}{\sqrt{2}} \quad \therefore \beta =\sqrt{2} \\\\ \Rightarrow \alpha =-1, \beta^2=2 \end{array} $$

Now, $\alpha$ and $\beta^2$ are roots of the quadratic equation is $x^2-x-2=0$