Circles
Chord Length
Grade 11
Question:
<p>The sum of the squares of the lengths of the chords intercepted on the circle, \(x^2 + y^2 = 16\), by the lines, \(x + y = n\), \(n \in N\), where \(N\) is the set of all natural numbers, is __________.</p>
Step-by-Step Solution
Key Concept: For a line at distance d from the center of a circle with radius r, the chord length is 2√(r² - d²). The distance from origin to line x + y = n is |n|/√2. Sum this over all natural numbers using the chord length formula.
<p><strong>Step 1:</strong> For circle x² + y² = 16, radius r = 4. For line x + y = n, distance from origin is d = |n|/√2.</p><p><strong>Step 2:</strong> Chord exists only when d ≤ r, so |n|/√2 ≤ 4, giving n ≤ 4√2 ≈ 5.66. Thus n ∈ {1, 2, 3, 4, 5}. Check n = 5: d = 5/√2 ≈ 3.54 < 4 ✓</p><p><strong>Step 3:</strong> Chord length L(n) = 2√(r² - d²) = 2√(16 - n²/2) = 2√((32 - n²)/2) = √(2(32 - n²)) = √(64 - 2n²)</p><p><strong>Step 4:</strong> Sum of squares of chord lengths:</p><p>Σ L(n)² = Σ(64 - 2n²) for n = 1 to 5</p><p>= (64 - 2) + (64 - 8) + (64 - 18) + (64 - 32) + (64 - 50)</p><p>= 62 + 56 + 46 + 32 + 14</p><p>= 210</p><p><strong>Alternative approach:</strong> Σ(64 - 2n²) = 5(64) - 2Σn² = 320 - 2(1 + 4 + 9 + 16 + 25) = 320 - 2(55) = 320 - 110 = 210</p><p>∴ Answer: <strong>210</strong></p>
Correct Answer: 210