Question:

<p><strong>Step 1: Find the center and radius of the first circle.</strong></p><p>Circle 1: x² + y² - 4x - 4y + 6 = 0</p><p>Rewrite: (x² - 4x + 4) + (y² - 4y + 4) + 6 - 4 - 4 = 0</p><p>(x - 2)² + (y - 2)² = 2</p><p>Center C₁ = (2, 2), radius r₁ = √2</p><p><strong>Step 2: Find the center and radius of the second circle.</strong></p><p>Circle 2: x² + y² - 10x - 10y + λ = 0</p><p>Rewrite: (x² - 10x + 25) + (y² - 10y + 25) + λ - 25 - 25 = 0</p><p>(x - 5)² + (y - 5)² = 50 - λ</p><p>Center C₂ = (5, 5), radius r₂ = √(50 - λ)</p><p><strong>Step 3: Calculate the distance between centers.</strong></p><p>d = √[(5-2)² + (5-2)²] = √(9 + 9) = √18 = 3√2</p><p><strong>Step 4: Apply the condition for exactly two common tangents.</strong></p><p>For exactly two tangents, the circles must intersect at two points:</p><p>|r₁ - r₂| < d < r₁ + r₂</p><p>|√2 - √(50 - λ)| < 3√2 < √2 + √(50 - λ)</p><p><strong>Step 5: Solve the right inequality.</strong></p><p>3√2 < √2 + √(50 - λ)</p><p>2√2 < √(50 - λ)</p><p>8 < 50 - λ</p><p>λ < 42</p><p><strong>Step 6: Solve the left inequality.</strong></p><p>|√2 - √(50 - λ)| < 3√2</p><p>This gives two cases:</p><p>Case 1: √2 - √(50 - λ) < 3√2 ⟹ -√(50 - λ) < 2√2 ⟹ √(50 - λ) > -2√2 (always true)</p><p>Case 2: √(50 - λ) - √2 < 3√2 ⟹ √(50 - λ) < 4√2 ⟹ 50 - λ < 32 ⟹ λ > 18</p><p><strong>Step 7: Verify that 50 - λ > 0.</strong></p><p>For r₂ to be real: 50 - λ > 0 ⟹ λ < 50</p><p><strong>Step 8: Combine all conditions.</strong></p><p>From Step 5: λ < 42</p><p>From Step 6: λ > 18</p><p>From Step 7: λ < 50 (automatically satisfied since 42 < 50)</p><p>Therefore: 18 < λ < 42</p><p>∴ Answer: D</p>