<p>The equation of circle is \((x-0)^2 + (y-2)^2 + \lambda x = 0\) which passes through the points \((-2, 4)\). The centre of the circle is:</p>
Step-by-Step Solution
Key Concept: Substitute the given point into the circle equation to find λ, then rewrite in standard form (x-h)² + (y-k)² = r² to identify the center (h,k).
<p><strong>Step 1:</strong> Substitute point (-2, 4) into the equation to find λ.</p><p>(-2-0)² + (4-2)² + λ(-2) = 0</p><p>4 + 4 - 2λ = 0</p><p>8 - 2λ = 0</p><p>λ = 4</p><p><strong>Step 2:</strong> Substitute λ = 4 back into the original equation.</p><p>x² + (y-2)² + 4x = 0</p><p>x² + 4x + (y-2)² = 0</p><p><strong>Step 3:</strong> Complete the square for x terms.</p><p>(x² + 4x + 4) - 4 + (y-2)² = 0</p><p>(x + 2)² + (y-2)² = 4</p><p><strong>Step 4:</strong> Identify the center from standard form (x-h)² + (y-k)² = r².</p><p>∴ Centre = (-2, 2)</p>
Correct Answer: A