<p>The centres of the circles \(x^2 + y^2 = 1\), \(x^2 + y^2 + 6x - 2y - 1 = 0\) and \(x^2 + y^2 - 12x + 4y = 1\) are</p>
Step-by-Step Solution
Key Concept: Convert each circle equation to standard form (x-h)² + (y-k)² = r² by completing the square to identify the center (h,k) directly from the coefficients.
<p><strong>Step 1:</strong> For circle <strong>x² + y² = 1</strong></p><p>This is already in standard form (x-0)² + (y-0)² = 1²</p><p>Center: <strong>C₁ = (0, 0)</strong></p><p><strong>Step 2:</strong> For circle <strong>x² + y² + 6x - 2y - 1 = 0</strong></p><p>Rearrange: (x² + 6x) + (y² - 2y) = 1</p><p>Complete the square: (x² + 6x + 9) + (y² - 2y + 1) = 1 + 9 + 1</p><p>(x + 3)² + (y - 1)² = 11</p><p>Center: <strong>C₂ = (-3, 1)</strong></p><p><strong>Step 3:</strong> For circle <strong>x² + y² - 12x + 4y = 1</strong></p><p>Rearrange: (x² - 12x) + (y² + 4y) = 1</p><p>Complete the square: (x² - 12x + 36) + (y² + 4y + 4) = 1 + 36 + 4</p><p>(x - 6)² + (y + 2)² = 41</p><p>Center: <strong>C₃ = (6, -2)</strong></p><p>∴ The centers are <strong>(0, 0), (-3, 1), and (6, -2)</strong></p>
Correct Answer: C