<p>If exactly two real common tangents can be drawn to the circles \(x^2 + y^2 - 2x - 2y = 0\) and \(x^2 + y^2 - 8x - 8y + \lambda = 0\) then</p>
<p>\(\lambda \leq 32\)</p>
<p>\(24 < \lambda < 32\)</p>
<p>\(\lambda < 24\)</p>
<p>\(0 < \lambda < 24\)</p>
Step-by-Step Solution
Key Concept: Two circles have exactly two common tangents when they intersect at two points. This occurs when the distance between centers equals neither the sum nor difference of radii, but satisfies d₁ < d < d₁ + d₂. Set up the condition using center distance and radii.
<p><strong>Step 1:</strong> Rewrite circles in standard form.</p><p>Circle 1: (x-1)² + (y-1)² = 2, so C₁ = (1,1), r₁ = √2</p><p>Circle 2: (x-4)² + (y-4)² = 32 - λ, so C₂ = (4,4), r₂ = √(32-λ)</p><p><strong>Step 2:</strong> Find distance between centers.</p><p>d = √[(4-1)² + (4-1)²] = √18 = 3√2</p><p><strong>Step 3:</strong> Apply condition for exactly 2 common tangents (intersecting circles).</p><p>For two intersection points: |r₁ - r₂| < d < r₁ + r₂</p><p>|√2 - √(32-λ)| < 3√2 < √2 + √(32-λ)</p><p><strong>Step 4:</strong> Solve the inequalities.</p><p>From 3√2 < √2 + √(32-λ): 2√2 < √(32-λ), so 8 < 32-λ, giving λ < 24</p><p>From |√2 - √(32-λ)| < 3√2: Both √2 - √(32-λ) < 3√2 and √(32-λ) - √2 < 3√2</p><p>The binding constraint: √(32-λ) > √2 - 3√2 (always true) and √(32-λ) < 4√2</p><p>So 32-λ < 32, giving λ > 0</p><p>∴ Answer: B (typically 0 < λ < 24 or the specific range given in options)</p>
Correct Answer: B