The sum of the squares of the lengths of the chords intercepted on the circle, , by the lines, , , where is the set of all natural numbers, is
- A.320
- B.105
- C.160
- D.210
Show answer and solution
Correct option: D
Solution
For the line x + y = n, the perpendicular distance from the origin (center of circle of radius 4) is d = n/√2.
The chord length is 2√(r² - d²) = 2√(16 - n²/2), so the square of the chord length is 4(16 - n²/2) = 64 - 2n².
For a chord to exist, we need 16 - n²/2 > 0, so n² < 32, meaning n ≤ 5 (since n ∈ N).
So n can be 1, 2, 3, 4, 5.
Sum = Σ(64 - 2n²) for n = 1 to 5
= 5(64) - 2(1 + 4 + 9 + 16 + 25)
= 320 - 2(55)
= 320 - 110
= 210
The chord length is 2√(r² - d²) = 2√(16 - n²/2), so the square of the chord length is 4(16 - n²/2) = 64 - 2n².
For a chord to exist, we need 16 - n²/2 > 0, so n² < 32, meaning n ≤ 5 (since n ∈ N).
So n can be 1, 2, 3, 4, 5.
Sum = Σ(64 - 2n²) for n = 1 to 5
= 5(64) - 2(1 + 4 + 9 + 16 + 25)
= 320 - 2(55)
= 320 - 110
= 210
[JEE Main 2019 · 8 April · Shift I]