QJEEM19Mad1cc: JEE Main 2019 Mathematics - Class 11 JEEM19Mad1cc Solution: The sum of the squares of the lengths of the chords intercepted …

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, ${l^2}$ = 62

For n = 2, ${l^2}$ = 56

For n = 3, ${l^2}$ = 46

For n = 4, ${l^2}$ = 32

For n = 5, ${l^2}$ = 14

$\therefore$ Sum of square of length of chords = 62 + 56 + 46 + 32 + 14 = 210