The equation of the circle, which is the mirror image of the circle, x² + y² - 2x = 0 in the line y = 3 – x is:

• A. x² + y² - 6x - 8y + 24 = 0
• B. x² + y² - 8x - 6y + 24 = 0
• C. x² + y² - 4x - 6y + 12 = 0
• D. x² + y² - 6x - 4y + 12 = 0

The given equation of the circle is x² + y² - 2x = 0. To find its mirror image across the line y = 3 - x, we first need to determine the center and radius of the original circle.

Step 1: Identify the Circle's Properties

The equation can be rewritten as:

x² - 2x + y² = 0

Completing the square for the x terms:

• (x - 1)² - 1 + y² = 0
• (x - 1)² + y² = 1

This shows that the circle has a center at (1, 0) and a radius of 1.

Step 2: Reflect the Center

To find the mirror image of the center (1, 0) across the line y = 3 - x, we can use the reflection formula. The line can be rewritten in standard form as x + y - 3 = 0.

Using the reflection formula, the reflected point (x', y') can be calculated. After performing the necessary calculations, the new center will be (4, 3).

Step 3: Write the New Circle's Equation

With the new center (4, 3) and the same radius of 1, the equation of the new circle is:

(x - 4)² + (y - 3)² = 1

Expanding this gives:

• x² - 8x + 16 + y² - 6y + 9 = 1
• x² + y² - 8x - 6y + 24 = 0

Final Answer

The equation of the circle, which is the mirror image of the original circle across the line y = 3 - x, is:

B. x² + y² - 8x - 6y + 24 = 0