Let a circle $C_1$ be obtained on rolling the circle $x^2+y^2-4x-6y+11=0$ upwards $4$ units on the tangent $\mathrm{T}$ to it at the point $\left(3,2\right)$. Let $C_2$ be the image of $C_1$ in $T$. Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$ respectively, and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium AMNB is:

• A

  $2\left(2+\sqrt{2}\right)$
• Correct answer
  B

  $4\left(1+\sqrt{2}\right)$
• C

  $3+2\sqrt{2}$
• D

  $2\left(1+\sqrt{2}\right)$

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