If a chord of the circle x^ 2 +y^ 2 =8 makes equal intercets of...

Since the chord makes equal intercepts of length a on the coordinate axes. So, its equation can be written as $\mathrm{x} \pm \mathrm{y}=\pm \mathrm{a}$. This line meets the given circle at two distinct points. So, length of the perpendicular from the centre (0,0) of the given circle must be less than the radius. i.e. $\left|\pm \frac{\mathrm{a}}{\sqrt{2}}\right|\lt \sqrt{8} \Rightarrow \mathrma^{2}\lt 16 \Rightarrow|\mathrm{a}|\lt 4$