question_answer

The sum of the squares of the lengths of the chords intercepted on the circle, \[{{x}^{2}}+{{y}^{2}}=16,\] by the lines, \[x+y=n,n\in N,\] where N is the set of all natural numbers, is : [JEE Main 8-4-2019 Morning]

A) 320                 

B) 160

C) 105                 

D) 210

Correct Answer: D

Solution :

\[p=\frac{n}{\sqrt{2}},\]but\[\frac{n}{\sqrt{2}}<4\Rightarrow n=1,2,3,4,5.\] Length of chord \[AB=2\sqrt{16-\frac{{{n}^{2}}}{2}}\] \[=\sqrt{64-2{{n}^{2}}}=\ell (say)\] For       \[n=1,{{\ell }^{2}}=62\]             \[n=2,{{\ell }^{2}}=56\]             \[n=3,{{\ell }^{2}}=46\]             \[n=4,{{\ell }^{2}}=32\]             \[n=5,{{\ell }^{2}}=14\] \[\therefore \]Required sum=62+56+46+32+14=210