Let a circle C 1 be obtained on rolling the circle x 2 + y 2 - 4 x - 6 y + 11 = 0 upwards 4 units on the…

We have,

$x^2+y^2-4x-6y+11=0$

$\Rightarrow {\left(x-2\right)}^2+{\left(y-3\right)}^2=2$

Centre$\equiv \left(2,3\right)$

Radius $r=\sqrt{2}$ units

Equation of tangent at $\left(3,2\right)$ is

$3x+2y-2\left(x+3\right)-3\left(y+2\right)+11=0$

$\Rightarrow x=y=1 ...\left(1\right)$

On rolling the given circle upward $4$ units on the tangent $\left(1\right)$, centre of the circle also moving $4$ unit upward, so

Centre of $C_1$ is

$=\left(2+2\sqrt{2},3+2\sqrt{2}\right)$

$A\left(2+2\sqrt{2},3+2\sqrt{2}\right)$

Centre of $C_2$ is

$B\left(4+2\sqrt{2},1+2\sqrt{2}\right)$

So, $M\left(2+2\sqrt{2},0\right)$ and $N\left(4+2\sqrt{2},0\right)$.

$\therefore$  area of trapezium

$=\frac{1}{2}\left(AM+BN\right)\times MN$

$=\frac{1}{2}\left(4+4\sqrt{2}\right)2=4\left(1+\sqrt{2}\right)$ sq. units