<p>Given two circles \((x-1)^2 + (y-3)^2 = l^2\) and \(x^2 + y^2 - 8x + 2y + 8 = 0\). If the two circles intersect in two distinct points, find the range of \(l\) (i.e., \(r\)):</p>
Step-by-Step Solution
Key Concept: Two circles intersect at two distinct points when the distance between centers d satisfies |r₁ - r₂| < d < r₁ + r₂. Convert the second circle to standard form to find both centers and radii, then set up the inequality.
<p><strong>Step 1:</strong> Identify Circle 1: Center C₁ = (1, 3), radius r₁ = l</p><p><strong>Step 2:</strong> Convert Circle 2 to standard form: x² + y² - 8x + 2y + 8 = 0<br/>Complete the square: (x - 4)² + (y + 1)² = 16 + 1 - 8 = 9<br/>Center C₂ = (4, -1), radius r₂ = 3</p><p><strong>Step 3:</strong> Find distance between centers:<br/>d = √[(4-1)² + (-1-3)²] = √[9 + 16] = √25 = 5</p><p><strong>Step 4:</strong> For two distinct intersection points, apply the condition:<br/>|r₁ - r₂| < d < r₁ + r₂<br/>|l - 3| < 5 < l + 3</p><p><strong>Step 5:</strong> Solve the compound inequality:<br/>From |l - 3| < 5: -5 < l - 3 < 5 ⟹ -2 < l < 8<br/>From 5 < l + 3: l > 2<br/>Combined: 2 < l < 8</p><p>∴ Answer: A (Range is 2 < l < 8 or l ∈ (2, 8))</p>
Correct Answer: A