If a chord of a circle $x^{2}+y^{2}=32$ makes equal intercepts of length $l$ on the co-ordinate axes, then
- $\ell <8$
- $\ell <16$
- $\ell >8$
- $\ell >16$
The line making equal intercepts l on axes is x/l + y/l = 1, or x + y = l. The distance from the center (0,0) to this line must be less than the radius (sqrt(32)). Distance = |l| / sqrt(1^2 + 1^2) = l / sqrt(2). So l / sqrt(2) < sqrt(32) => l < sqrt(64) => l < 8.
A line making equal intercepts of length l on the coordinate axes has the equation x + y = l or x - y = l. The perpendicular distance from the origin to this chord cannot exceed the radius of the circle, which is the square root of 32, or 4 times the square root of 2. Equating the perpendicular distance p, given by l divided by the square root of 2, to the maximum possible distance of 4 times the square root of 2 yields l = 8. Because the chord must lie inside the circle and cannot pass through the ends of the diameter to maintain an exact intercept without being tangent, its distance to the center must be strictly less than the radius, meaning l is less than 8.