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If a chord of the circle $x^{2}+y^{2}=8$ makes equal intercets of length a on the coordinate axes, then
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$|\mathrm{a}|<4$
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. $\left|\pm \frac{\mathrm{a}}{\sqrt{2}}\right|<\sqrt{8} \Rightarrow \mathrm{a}^{2}<16 \Rightarrow|\mathrm{a}|<4$
i.e. $\left|\pm \frac{\mathrm{a}}{\sqrt{2}}\right|<\sqrt{8} \Rightarrow \mathrm{a}^{2}<16 \Rightarrow|\mathrm{a}|<4$
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