If the circle `x^(2) + y^(2) -4x - 8y + 16 =0` rolls up the tangent to it at `(2 + sqrt(3), 3)` by 2 units (assumes x-axis as horizontal), then the centre of the circle in the new position is

Options

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

Answer

To solve the problem, we need to follow these steps:

### Step 1: Rewrite the equation of the circle in standard form
The given equation of the circle is:
\[ x^2 + y^2 - 4x - 8y + 16 = 0 \]

We can rearrange this equation to find the center and radius of the circle. To do this, we will complete the square for both \(x\) and \(y\).

1. For \(x\):
\[ x^2 - 4x \rightarrow (x - 2)^2 - 4 \]

2. For \(y\):
\[ y^2 - 8y \rightarrow (y - 4)^2 - 16 \]

Now substituting back into the equation:
\[ (x - 2)^2 - 4 + (y - 4)^2 - 16 + 16 = 0 \]
This simplifies to:
\[ (x - 2)^2 + (y - 4)^2 = 4 \]

### Step 2: Identify the center and radius of the circle
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \(A\) of the circle: \( (h, k) = (2, 4) \)
- Radius \(r = 2\)

### Step 3: Find the equation of the tangent at the given point
The point given is \(P(2 + \sqrt{3}, 3)\). The equation of the tangent to the circle at point \(P\) can be derived using the formula:
\[ xx_1 + yy_1 - g(x + x_1) - f(y + y_1) + c = 0 \]
Where \(g = -2\), \(f = -4\), and \(c = 16\).

Substituting \(x_1 = 2 + \sqrt{3}\) and \(y_1 = 3\):
\[ x(2 + \sqrt{3}) + y(3) + 2(x + (2 + \sqrt{3})) + 4(y + 3) + 16 = 0 \]

After simplification, we find:
\[ \sqrt{3}x - y - 2\sqrt{3} = 0 \]

### Step 4: Determine the slope of the tangent
From the tangent equation \(\sqrt{3}x - y - 2\sqrt{3} = 0\), we can express \(y\) in terms of \(x\):
\[ y = \sqrt{3}x - 2\sqrt{3} \]
The slope \(m\) of the tangent line is \(\sqrt{3}\).

### Step 5: Calculate the new position of the center after rolling
The angle \(\theta\) that the tangent makes with the x-axis can be found using:
\[ \tan \theta = \sqrt{3} \implies \theta = 60^\circ \]

Now, the center \(B\) of the circle after rolling up the tangent by 2 units can be calculated using:
\[ B = (x_A + k \cos \theta, y_A + k \sin \theta) \]
Where \(k = 2\) (the distance rolled), and \(\cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).

Substituting the values:
\[ B = (2 + 2 \cdot \frac{1}{2}, 4 + 2 \cdot \frac{\sqrt{3}}{2}) = (2 + 1, 4 + \sqrt{3}) = (3, 4 + \sqrt{3}) \]

### Final Answer
The center of the circle in the new position is:
\[ \boxed{(3, 4 + \sqrt{3})} \]