Find the equation of the image of the circle `x^(2)+y^(2)-6x-4y+12=0` by the line mirror x+y-1=0

Options

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

Answer

To find the equation of the image of the circle \(x^{2} + y^{2} - 6x - 4y + 12 = 0\) with respect to the line mirror \(x + y - 1 = 0\), we will follow these steps:

### Step 1: Identify the center and radius of the circle
The given equation of the circle can be rewritten in standard form. We start by completing the square.

The equation is:
\[ x^{2} - 6x + y^{2} - 4y + 12 = 0 \]

Completing the square for \(x\):
\[ x^{2} - 6x = (x - 3)^{2} - 9 \]

Completing the square for \(y\):
\[ y^{2} - 4y = (y - 2)^{2} - 4 \]

Substituting back into the equation:
\[ (x - 3)^{2} - 9 + (y - 2)^{2} - 4 + 12 = 0 \]
\[ (x - 3)^{2} + (y - 2)^{2} - 1 = 0 \]
Thus, we have:
\[ (x - 3)^{2} + (y - 2)^{2} = 1 \]

From this, we can identify the center of the circle as \(C(3, 2)\) and the radius \(r = 1\).

### Step 2: Find the image of the center with respect to the line
To find the image of the center \(C(3, 2)\) with respect to the line \(x + y - 1 = 0\), we can use the formula for the reflection of a point across a line.

The line can be rewritten in the form \(Ax + By + C = 0\) where \(A = 1\), \(B = 1\), and \(C = -1\).

The formula for the reflection of a point \((x_0, y_0)\) across the line \(Ax + By + C = 0\) is given by:
\[ \left( x', y' \right) = \left( x_0 - \frac{2A(Ax_0 + By_0 + C)}{A^2 + B^2}, y_0 - \frac{2B(Ax_0 + By_0 + C)}{A^2 + B^2} \right) \]

Substituting \(C(3, 2)\) into the formula:
\[ Ax_0 + By_0 + C = 1 \cdot 3 + 1 \cdot 2 - 1 = 4 \]
\[ A^2 + B^2 = 1^2 + 1^2 = 2 \]

Now substituting into the reflection formula:
\[ x' = 3 - \frac{2 \cdot 1 \cdot 4}{2} = 3 - 4 = -1 \]
\[ y' = 2 - \frac{2 \cdot 1 \cdot 4}{2} = 2 - 4 = -2 \]

Thus, the image of the center is \(C'(-1, -2)\).

### Step 3: Write the equation of the image circle
The radius remains the same, \(r = 1\). Therefore, the equation of the image circle with center \(C'(-1, -2)\) and radius \(1\) is:
\[ (x + 1)^{2} + (y + 2)^{2} = 1 \]

Expanding this gives:
\[ (x^{2} + 2x + 1) + (y^{2} + 4y + 4) = 1 \]
\[ x^{2} + y^{2} + 2x + 4y + 5 = 1 \]
\[ x^{2} + y^{2} + 2x + 4y + 4 = 0 \]

### Final Answer
The equation of the image of the circle is:
\[ x^{2} + y^{2} + 2x + 4y + 4 = 0 \]