Without changing the direction of the axes, the origin is transferred to the point (2,3). Then the equation `x^(2)+y^(2)-4x-6y+9=0` changes to

Options

A. `x^(2)+y^(2)+4=0`
B. `x^(2)+y^(2)=4`
C. `x^(2)+y^(2)-8x-12y+48=0`
D. `x^(2)+y^(2)=9`

Answer

To solve the problem, we need to transfer the origin of the coordinate system from (0, 0) to (2, 3) without changing the direction of the axes. This means we will be using a new set of coordinates (x', y') instead of the original coordinates (x, y).

### Step-by-Step Solution:

1. **Identify the transformation**:
The new coordinates (x', y') are related to the old coordinates (x, y) by the equations:
\[ x = x' + 2 \]
\[ y = y' + 3 \]

2. **Substitute the transformations into the original equation**:
The original equation is:
\[ x^2 + y^2 - 4x - 6y + 9 = 0 \]
We will replace \(x\) and \(y\) with \(x' + 2\) and \(y' + 3\) respectively.

Substituting these values gives:
\[ (x' + 2)^2 + (y' + 3)^2 - 4(x' + 2) - 6(y' + 3) + 9 = 0 \]

3. **Expand the equation**:
Expanding the squares and simplifying:
\[ (x'^2 + 4x' + 4) + (y'^2 + 6y' + 9) - (4x' + 8) - (6y' + 18) + 9 = 0 \]
Combine like terms:
\[ x'^2 + y'^2 + 4x' + 6y' + 4 + 9 - 4x' - 8 - 6y' - 18 + 9 = 0 \]
Simplifying further:
\[ x'^2 + y'^2 + (4x' - 4x') + (6y' - 6y') + (4 + 9 - 8 - 18 + 9) = 0 \]
This reduces to:
\[ x'^2 + y'^2 - 4 = 0 \]

4. **Rearranging the equation**:
We can rearrange this to:
\[ x'^2 + y'^2 = 4 \]

### Final Result:
The transformed equation is:
\[ x'^2 + y'^2 = 4 \]