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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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S solutionqc

$\\x^{2}+y^{2}=16\\\; \; \; x+y=n$

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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