The sum of the square of the length of the chords intercepted by the line x + y = n, n∈N on the circle x2 + y2 = 4 is p then p11is equal to

AB2=4AM244−n22=28−n2,n∈N⇒n=1 or 2Hencep=28−12+8−22=22 ∴p11=2

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The sum of the square of the length of the chords intercepted by the line x + y = n, $n \in N$ on the circle x² + y² = 4 is p then $\frac{p}{11}$ is equal to

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answer is 2.

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Detailed Solution

$\begin{array}{l} {AB}^{2} = 4 {AM}^{2} \\ 4 (4 - \frac{n^{2}}{2}) = 2 (8 - n^{2}), n \in N \Rightarrow n = 1 or 2 \end{array}$
Hence
$p = 2 (8 - 1^{2} + 8 - 2^{2}) = 22 ∴ \frac{p}{11} = 2$

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