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 T to it at the point `(3,2)`.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:

Options

A. `2(2+sqrt(2))`
B. `2(1+sqrt(2))`
C. `4(1+sqrt(2))`
D. `3+2sqrt(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the center and radius of the original circle \( C_1 \)

The equation of the circle is given as:
\[ x^2 + y^2 - 4x - 6y + 11 = 0 \]
We can rewrite this in standard form by completing the square.

1. Rearranging the terms:
\[ (x^2 - 4x) + (y^2 - 6y) + 11 = 0 \]

2. Completing the square:
- For \( x \): \( x^2 - 4x = (x - 2)^2 - 4 \)
- For \( y \): \( y^2 - 6y = (y - 3)^2 - 9 \)

3. Substituting back:
\[ (x - 2)^2 - 4 + (y - 3)^2 - 9 + 11 = 0 \]
\[ (x - 2)^2 + (y - 3)^2 - 2 = 0 \]
\[ (x - 2)^2 + (y - 3)^2 = 2 \]

Thus, the center \( A \) of circle \( C_1 \) is \( (2, 3) \) and the radius \( r \) is \( \sqrt{2} \).

### Step 2: Find the tangent line at the point \( (3, 2) \)

The slope of the radius at point \( (3, 2) \) can be calculated as follows:
- The center \( A \) is \( (2, 3) \).
- The slope of the line from \( A \) to \( (3, 2) \) is:
\[ \text{slope} = \frac{2 - 3}{3 - 2} = -1 \]

The tangent line at this point will have a slope that is the negative reciprocal, which is \( 1 \). The equation of the tangent line can be written using point-slope form:
\[ y - 2 = 1(x - 3) \]
Simplifying gives:
\[ y = x - 1 \]

### Step 3: Determine the new center after rolling the circle upwards by 4 units

Since the circle rolls upwards along the tangent line \( y = x - 1 \), we add 4 to the y-coordinate of the center \( A \):
- New center \( A' \):
\[ A' = (2, 3 + 4) = (2, 7) \]

### Step 4: Find the image of \( C_1 \) in the tangent line \( T \)

The image center \( B \) can be found by reflecting \( A' \) across the line \( y = x - 1 \).

1. The line \( y = x - 1 \) can be rewritten in standard form as:
\[ x - y - 1 = 0 \]

2. To find the reflection of point \( A' \) across this line, we can use the formula for reflection across a line \( Ax + By + C = 0 \):
\[ \text{Reflection} = \left( x - \frac{2A(Ax + By + C)}{A^2 + B^2}, y - \frac{2B(Ax + By + C)}{A^2 + B^2} \right) \]
For \( A = 1, B = -1, C = -1 \):
\[ Ax + By + C = 2 - 7 - 1 = -6 \]
Calculating the reflection:
\[ B = \left( 2 - \frac{2 \cdot 1 \cdot (-6)}{1^2 + (-1)^2}, 7 - \frac{2 \cdot (-1) \cdot (-6)}{1^2 + (-1)^2} \right) \]
\[ = \left( 2 + \frac{12}{2}, 7 - \frac{12}{2} \right) = (2 + 6, 7 - 6) = (8, 1) \]

### Step 5: Find the feet of the perpendiculars \( M \) and \( N \)

- The foot of the perpendicular from \( A' = (2, 7) \) to the x-axis is:
\[ M = (2, 0) \]

- The foot of the perpendicular from \( B = (8, 1) \) to the x-axis is:
\[ N = (8, 0) \]

### Step 6: Calculate the area of trapezium \( AMNB \)

The area \( A \) of trapezium \( AMNB \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h \]
Where:
- \( b_1 = AM = 7 \) (height from \( A' \) to the x-axis)
- \( b_2 = BN = 1 \) (height from \( B \) to the x-axis)
- \( h = |x_A - x_B| = |2 - 8| = 6 \)

Calculating the area:
\[ \text{Area} = \frac{1}{2} \times (7 + 1) \times 6 = \frac{1}{2} \times 8 \times 6 = 24 \]

### Final Answer
The area of trapezium \( AMNB \) is \( 24 \).