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Solve this problem - Co-ordinate geometry - JEE Main-6

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\epsilon N, where N is the set of all natural numbers, is :

  • Option 1)

    105

  • Option 2)

    210

  • Option 3)

    160

  • Option 4)

    320

 
Answers (1)
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\\x^{2}+y^{2}=16\\\; \; \; x+y=n

then length of perpendicular from centre (0,0) to line x+y=n = \left | \frac{0+0-n}{\sqrt{1^{2}+1^{2}}} \right |=\frac{n}{\sqrt{2}}

length of intercepts =2\sqrt{4^{2}-\frac{n^2}{(\sqrt{2})^{2}}}=2\sqrt{16-\frac{n^{2}}{2}}

=\sqrt{64-2n^{2}}

Possible value of n are = 1,2,3,4,5

Sum of squares of length =\sum_{i=1}^{5}(\sqrt{64-2x^{2}})^{2}

=\sum_{i=1}^{5}(64-2x^{2})

=64\times5-2\; \; \; \sum_{i=1}^{5}n^{2}

=64\times5-2\; \; \; \frac{n(n+1)(2n+1)}{6}

=320-2\; \frac{5\times6\times11}{6}

=320-110

=210


Option 1)

105

Option 2)

210

Option 3)

160

Option 4)

320

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