JEE Main 2019 — Circle Question with 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 = $$\left|\frac{0+0-n}{\sqrt{1^2+1^2}}\right|=\frac{n}{\sqrt{2}}$$

Radius of the circle = 4

From the picture you can see,

$$\frac{n}{\sqrt{2}}$$ < 4

$$\Rightarrow$$ n < 5.65

$$\therefore$$ Possible value of n = 1, 2, 3, 4, 5

Length of chord AB = $$2\sqrt{{\left(4\right)}^2-{\left(\frac{n}{\sqrt{2}}\right)}^2}$$

= $$2\sqrt{16-\frac{n^2}{2}}$$

= $$\sqrt{64-2n^2}$$ = $$l$$

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