P and Q are the ends of a diameter of the circle $\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2\left(\mathrm{a}>\frac{1}{\sqrt{2}}\right) \cdot s$ and $t$ are the lengths of the perpendiculars drawn from $P$ and $Q$ onto the line $x+y=1$ respectively. When the product $s t$ is maximum, the greater value among $s, t$ is

• A

  $a+\sqrt{2}$
• Correct answer
  B

  $a+\frac{1}{\sqrt{2}}$
• C

  $a-\frac{1}{\sqrt{2}}$
• D

  $a-\sqrt{2}$

Hence, the correct answer is option: 'B'.