Question:

Step 1: Determine the center and radius of the initial circle and the equation of the tangent. The equation of the given circle is $x^2+y^2-4x-6y+11=0$. We complete the square to find its standard form: $$ (x^2-4x+4) + (y^2-6y+9) + 11 - 4 - 9 = 0 $$ $$ (x-2)^2 + (y-3)^2 = 2 $$ The center of the initial circle, $C_0$, is $(2,3)$ and its radius, $r$, is $\sqrt{2}$. The equation of the tangent $T$ to the circle $(x-h)^2+(y-k)^2=r^2$ at a point $(x_1, y_1)$ is $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$. For $C_0=(2,3)$, $r^2=2$, and point $(3,2)$: $$ (x-2)(3-2) + (y-3)(2-3) = 2 $$ $$ (x-2)(1) + (y-3)(-1) = 2 $$ $$ x-2-y+3 = 2 $$ $$ x-y+1 = 2 $$ Thus, the equation of the tangent $T$ is $x-y-1=0$. Step 2: Find the coordinates of the center $A$ of circle $C_1$. The phrase "rolling the circle upwards 4 units on the tangent $T$ at $(3,2)$" implies that the original point of tangency $(3,2)$ moves along the tangent line $T$ by 4 units in the "upward" direction. The "upward" direction along the tangent $x-y-1=0$ (which has a positive slope $m=1$) corresponds to increasing both $x$ and $y$ coordinates. The unit vector along $T$ in the upward direction is $\hat{t} = \left(\frac{1}{\sqrt{1^2+1^2}}, \frac{1}{\sqrt{1^2+1^2}}\right) = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. The new point of tangency, $P'$, is obtained by moving 4 units from $(3,2)$ along $\hat{t}$: $$ P' = (3,2) + 4\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = (3,2) + (2\sqrt{2}, 2\sqrt{2}) = (3+2\sqrt{2}, 2+2\sqrt{2}) $$ The center $A$ of circle $C_1$ is at a distance $r=\sqrt{2}$ from $P'$ along the normal to $T$, in the direction away from the x-axis or towards the "upward" side of $T$. The normal vector to $T: x-y-1=0$ is $(1,-1)$ or $(-1,1)$. The original center $C_0=(2,3)$ satisfies $2-3-1 = -2 < 0$, which means it lies on the "upward" side of the tangent line. Therefore, the unit normal vector pointing towards this "upward" side is $\hat{n} = \frac{(-1,1)}{\sqrt{1^2+(-1)^2}} = \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$. The center $A$ of $C_1$ is: $$ A = P' + r \hat{n} = (3+2\sqrt{2}, 2+2\sqrt{2}) + \sqrt{2} \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) $$ $$ A = (3+2\sqrt{2}, 2+2\sqrt{2}) + (-1,1) = (2+2\sqrt{2}, 3+2\sqrt{2}) $$ Step 3: Determine the coordinates of the center $B$ of circle $C_2$. Circle $C_2$ is the image of $C_1$ in the tangent $T: x-y-1=0$. Let $A=(x_A, y_A)=(2+2\sqrt{2}, 3+2\sqrt{2})$ and $B=(x_B, y_B)$ be its image. The reflection formula for a point $(x,y)$ across the line $ax+by+c=0$ is: $$ \frac{x'-x}{a} = \frac{y'-y}{b} = -2\frac{ax+by+c}{a^2+b^2} $$ Here, $a=1, b=-1, c=-1$. Substitute the coordinates of $A$: $$ 1 \cdot (2+2\sqrt{2}) - 1 \cdot (3+2\sqrt{2}) - 1 = 2+2\sqrt{2}-3-2\sqrt{2}-1 = -2 $$ Now apply the reflection formula for $A(2+2\sqrt{2}, 3+2\sqrt{2})$ to find $B(x_B, y_B)$: $$ \frac{x_B - (2+2\sqrt{2})}{1} = \frac{y_B - (3+2\sqrt{2})}{-1} = -2 \frac{-2}{1^2+(-1)^2} $$ $$ \frac{x_B - (2+2\sqrt{2})}{1} = \frac{y_B - (3+2\sqrt{2})}{-1} = -2 \frac{-2}{2} = 2 $$ From the first part: $$ x_B - (2+2\sqrt{2}) = 2 \implies x_B = 4+2\sqrt{2} $$ From the second part: $$ y_B - (3+2\sqrt{2}) = -2 \implies y_B = 1+2\sqrt{2} $$ So, the center $B$ of circle $C_2$ is $(4+2\sqrt{2}, 1+2\sqrt{2})$. Step 4: Calculate the area of the trapezium AMNB. $A=(x_A, y_A) = (2+2\sqrt{2}, 3+2\sqrt{2})$ $B=(x_B, y_B) = (4+2\sqrt{2}, 1+2\sqrt{2})$ $M$ is the foot of the perpendicular from $A$ to the x-axis, so $M=(x_A, 0) = (2+2\sqrt{2}, 0)$. $N$ is the foot of the perpendicular from $B$ to the x-axis, so $N=(x_B, 0) = (4+2\sqrt{2}, 0)$. The trapezium AMNB has parallel sides $AM$ and $BN$ (both vertical). Length of side $AM = y_A = 3+2\sqrt{2}$. Length of side $BN = y_B = 1+2\sqrt{2}$. The height of the trapezium is the horizontal distance between $M$ and $N$: Height $h = |x_B - x_A| = |(4+2\sqrt{2}) - (2+2\sqrt{2})| = |2| = 2$. The area of a trapezium is given by $\frac{1}{2}(\text{sum of parallel sides}) \times \text{height}$. Area of trapezium AMNB $= \frac{1}{2} (AM + BN) \cdot h$ $$ \text{Area} = \frac{1}{2} ((3+2\sqrt{2}) + (1+2\sqrt{2})) \cdot 2 $$ $$ \text{Area} = (4+4\sqrt{2}) = 4(1+\sqrt{2}) $$ The area of the trapezium AMNB is $4(1+\sqrt{2})$. This matches Option 2. The final answer is $\boxed{4(1+\sqrt{2})}$.