<p>For any \(\lambda \in \mathbb{R}\), the locus of \(x^2 + y^2 - 2\lambda x - 2\lambda y + \lambda^2 = 0\) touches the line</p>
Step-by-Step Solution
Key Concept: Rewrite the equation as a family of circles by completing the square and identify the locus of centers. The envelope of this family of circles (the line they all touch) is found by eliminating λ from the circle equation and its derivative with respect to λ.
<p><strong>Step 1:</strong> Complete the square in the given equation:</p><p>x² + y² - 2λx - 2λy + λ² = 0</p><p>(x² - 2λx + λ²) + (y² - 2λy + λ²) - λ² = 0</p><p>(x - λ)² + (y - λ)² = λ²</p><p><strong>Step 2:</strong> This represents a family of circles with center (λ, λ) and radius |λ|. As λ varies, the center moves along the line y = x.</p><p><strong>Step 3:</strong> To find the envelope (common tangent), differentiate the original equation with respect to λ:</p><p>-2x - 2y + 2λ = 0</p><p>λ = x + y</p><p><strong>Step 4:</strong> Substitute λ = x + y back into the original equation:</p><p>x² + y² - 2(x + y)x - 2(x + y)y + (x + y)² = 0</p><p>x² + y² - 2x² - 2xy - 2xy - 2y² + x² + 2xy + y² = 0</p><p>-2xy = 0</p><p>Therefore, the locus touches the lines xy = 0, which means <strong>x = 0 (y-axis) or y = 0 (x-axis)</strong></p><p>∴ Answer: B</p>
Correct Answer: B