The sum of the squares of the lengths of the chords
intercepted on the circle, x2 + y2 = 16, by the lines,
x + y = n, n \[ \in \] N, where N is the set of all natural
numbers, is :
Solutions · 1
Official solution
Let the chord x + y = n cuts the circle x2 +
y2 = 16 at A and B.
Length of perpendicular from
O on AB,
OM = \[\left| {{{0 + 0 - n} \over {\sqrt {{1^2} + {1^2}} }}} \right| = {n \over {\sqrt 2 }}\]
Radius of the circle = 4
From the picture you can see,
\[{n \over {\sqrt 2 }}\] < 4
\[ \Rightarrow \] n < 5.65
\[ \therefore \] Possible value of n = 1, 2, 3, 4, 5
Length of chord AB = \[2\sqrt {{{\left( 4 \right)}^2} - {{\left( {{n \over {\sqrt 2 }}} \right)}^2}} \]
= \[2\sqrt {16 - {{{n^2}} \over 2}} \]
= \[\sqrt {64 - 2{n^2}} \] = l
For n = 1, l2 = 62
For n = 2, l2 = 56
For n = 3, l2 = 46
For n = 4, l2 = 32
For n = 5, l2 = 14
\[ \therefore \] Sum of square of length of chords
= 62 + 56 + 46 + 32 + 14 = 210