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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2019 · 8 April · Morning Shift · Mathematics · Circle

The sum of the squares of the lengths of the chords intercepted on...

The sum of the squares of the lengths of the chords intercepted on the circle, x2+y2=16,{x}^{2} + {y}^{2} = 16, by the lines, x+y=n,n∈N,x + y = n, n \in N, where NN is the set of all natural numbers is: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

Options

  1. A210210
  2. B105105
  3. C320320
  4. D160160

Correct answer

A. 210210

Step-by-step solution

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

The length of the perpendicular from centre (0,0)\left(0, 0\right) to the line x+y=nx + y = n is \left|\dfrac{0 + 0 - n}{\sqrt{{1}^{2} + {I}^{2}}}\right| = \dfrac{n}{\sqrt{2}}</p><p>Lengthofintercept</p><p>Length of intercept= 2\sqrt{{4}^{2} - {\left(\dfrac{n}{\sqrt{2}}\right)}^{2}} = \sqrt{64 - 2{n}^{2}}</p><p></p><p>⇒PossiblevaluesofPossible values ofnareare1, 2, \ldots . 5</p><p></p><p>⇒Sumofthesquareoflengthsofintercepts</p><p>Sum of the square of lengths of intercepts</p><p>= \mathop{Σ}_{n = 1}^{5}\left(64 - 2{n}^{2}\right) = 64 \times 5 - 2\mathop{Σ}_{n = 1}^{5}{n}^{2}</p><p></p><p>= 320 - 2\mathop{Σ}_{n = 1}^{5}{n}^{2} = 320 - 110 = 210

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