The equation of a circle, which is mirror change of the circle `x ^(2) + y ^(2) - 2x =0,` in the line,` y =3 -x` is

Options

A. `x ^(2) +y ^(2) - 6x - 4y + 12=0`
B. `x ^(2) + y ^(2) - 6x - 8y + 24=0`
C. `x ^(2) + y ^(2) - 8x - 6y + 24=0`
D. `x ^(2) +y ^(2) - 4x - 6y + 12=0`

Answer

To find the equation of the circle that is the mirror image of the circle given by the equation \( x^2 + y^2 - 2x = 0 \) in the line \( y = 3 - x \), we will follow these steps:

### Step 1: Identify the center and radius of the original circle
The given equation of the circle is:
\[ x^2 + y^2 - 2x = 0 \]
We can rewrite this as:
\[ x^2 - 2x + y^2 = 0 \]
Completing the square for the \( x \) terms:
\[ (x - 1)^2 - 1 + y^2 = 0 \]
\[ (x - 1)^2 + y^2 = 1 \]
From this, we can see that the center \( C_1 \) of the circle is \( (1, 0) \) and the radius \( r \) is \( 1 \).

### Step 2: Write the equation of the line in standard form
The line given is:
\[ y = 3 - x \]
We can rewrite this in standard form:
\[ x + y - 3 = 0 \]
Here, \( a = 1 \), \( b = 1 \), and \( c = -3 \).

### Step 3: Find the mirror image of the center of the circle
To find the mirror image of the point \( (1, 0) \) across the line \( x + y - 3 = 0 \), we use the formula for the mirror image of a point \( (x_1, y_1) \) across the line \( ax + by + c = 0 \):
\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{2(ax_1 + by_1 + c)}{a^2 + b^2} \]
Substituting \( (x_1, y_1) = (1, 0) \):
\[ \frac{x - 1}{1} = \frac{y - 0}{1} = -\frac{2(1 \cdot 1 + 1 \cdot 0 - 3)}{1^2 + 1^2} \]
Calculating the right-hand side:
\[ = -\frac{2(1 - 3)}{1 + 1} = -\frac{2(-2)}{2} = 2 \]
Thus, we have:
\[ x - 1 = 2 \quad \text{and} \quad y = 2 \]
From \( x - 1 = 2 \), we get \( x = 3 \). Therefore, the coordinates of the mirror image center \( C_2 \) are \( (3, 2) \).

### Step 4: Write the equation of the new circle
The new circle will have the center \( (3, 2) \) and the same radius \( r = 1 \). The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 3 \), \( k = 2 \), and \( r = 1 \):
\[ (x - 3)^2 + (y - 2)^2 = 1 \]

### Step 5: Expand the equation
Expanding this, we get:
\[ (x^2 - 6x + 9) + (y^2 - 4y + 4) = 1 \]
Combining terms:
\[ x^2 + y^2 - 6x - 4y + 13 = 1 \]
Subtracting 1 from both sides:
\[ x^2 + y^2 - 6x - 4y + 12 = 0 \]

### Final Answer
The equation of the circle which is the mirror image of the given circle in the line \( y = 3 - x \) is:
\[ x^2 + y^2 - 6x - 4y + 12 = 0 \]