,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$

Also $P$ and $Q$ are end points of diameter.

Let $P \equiv(a \cos \theta, a \sin \theta)$

$$ Q \equiv(-a \cos \theta,-a \sin \theta) $$

According to the question,

$$ \begin{aligned} s & =\frac{|a \cos \theta+a \sin \theta-1|}{\sqrt{2}} \\ & =\frac{|a(\cos \theta+\sin \theta)-1|}{\sqrt{2}} \\ \text { and } t & =\frac{|-a(\cos \theta+\sin \theta)-1|}{\sqrt{2}}=1 \\ \text { Now, st } & =\frac{\left|1-a^2(\cos \theta+\sin \theta)^2\right|}{2} \\ & =\frac{\left|1-a^2(1+\sin 2 \theta)\right|}{2} \end{aligned} $$

So, $(s t)$ will be maximum, if $(1+\sin 2 \theta)$ minimum,

$$ \begin{aligned} & \because(\sin 2 \theta)_{\min }=-1(\text { i.e. } \sin 2 \theta \geq-1) \\ & \Rightarrow 1+\sin 2 \theta \geq 0 \therefore(1+\sin 2 \theta)_{\min }=0 \\ & \therefore\left(a^2(1+\sin 2 \theta)\right)_{\min }=0 \text { and hence, } s t=\frac{1}{2} \end{aligned} $$

Also, $\left(a^2(1+\sin 2 \theta)\right)_{\max }=2 a^2$

$$ \Rightarrow \quad s t=\frac{\left|1-2 a^2\right|}{2}>\frac{1}{2} \text { for } a>\frac{1}{\sqrt{2}} $$

So, $(s t)_{\text {max }}$ occurs when

$$ \begin{aligned} & a^2(1+\sin 2 \theta)=2 a^2 \\ & \Rightarrow \quad 1+\sin 2 \theta=2\left(\because a>\frac{1}{\sqrt{2}} \therefore|a|=a\right) \\ & \Rightarrow \quad \cos \theta+\sin \theta=\sqrt{2} \\ & \Rightarrow \quad \theta=45^{\circ} \end{aligned} $$

$$ \begin{aligned} \therefore & & s & =\frac{|\sqrt{2} a-1|}{\sqrt{2}}, \text { and } t=\frac{\mid \sqrt{2 a+1 \mid}}{\sqrt{2}} \\ \therefore & & a & >\frac{1}{\sqrt{2}} \\ \Rightarrow & & \sqrt{2} a & >1 \\ \therefore & & s & =\frac{\sqrt{2} a-1}{\sqrt{2}}, \\ & & t & =\frac{\sqrt{2} a+1}{\sqrt{2}}=a+\frac{1}{\sqrt{2}} \end{aligned} $$

Clearly, $t>s$

$\therefore \quad\left(a+\frac{1}{\sqrt{2}}\right)$ is the final answer.

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