<p>If the radius of the circle \((x-1)^2 + (y-2)^2 = 1\) and \((x-7)^2 + (y-10)^2 = 4\) are increasing uniformly w.r.t. times as 0.3 and 0.4 unit/sec, then they will touch each other at \(t\) equal to</p>
Step-by-Step Solution
Key Concept: Two circles touch when the distance between their centers equals the sum of their radii (external tangency) or absolute difference (internal tangency). Set up the distance equation with time-dependent radii and solve for t.
<p><strong>Step 1: Identify the circles</strong></p><p>Circle 1: Center C₁ = (1, 2), Initial radius r₁(0) = 1</p><p>Circle 2: Center C₂ = (7, 10), Initial radius r₂(0) = 2</p><p><strong>Step 2: Find distance between centers</strong></p><p>d = √[(7-1)² + (10-2)²] = √[36 + 64] = √100 = 10</p><p><strong>Step 3: Set up radius functions with time</strong></p><p>r₁(t) = 1 + 0.3t</p><p>r₂(t) = 2 + 0.4t</p><p><strong>Step 4: Apply tangency condition</strong></p><p>For external tangency: d = r₁(t) + r₂(t)</p><p>10 = (1 + 0.3t) + (2 + 0.4t)</p><p>10 = 3 + 0.7t</p><p>0.7t = 7</p><p>t = 10 seconds</p><p><strong>Step 5: Verify</strong></p><p>At t = 10: r₁ = 1 + 3 = 4, r₂ = 2 + 4 = 6</p><p>Sum = 4 + 6 = 10 = d ✓</p><p>∴ Answer: B (t = 10 seconds)</p>
Correct Answer: B