Question

The chord of the circle $x^2+y^2=32$ makes equal intercepts of length $l$ on the co-ordinate axes then $|l|<8$.

Answer 1

Any chord making equal intercepts on co-ordinate axes is $x\pm y=\pm l$. Since it is a chord hence perpendicular from centre $(0,0)$ must be less than the radius $4\sqrt{2}$.
$\therefore |\pm \frac{l}{\sqrt{2}}|<4\sqrt{2}\therefore |l|<8$.