lf the line $\displaystyle \frac{x}{\mathrm{a}}+\frac{y}{b}=1$ intersects the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=\mathrm{c}^{2}$, then the length of the chord so formed is $(\sqrt{2}) \mathrm{c}$, then
The distance from the origin to the line x/a + y/b = 1 is d = 1 / sqrt(1/a^2 + 1/b^2). For a chord of length sqrt(2)c in a circle of radius c, the distance from the center is sqrt(c^2 - (sqrt(2)c/2)^2) = c/sqrt(2). Thus, 1/(1/a^2 + 1/b^2) = c^2/2, which simplifies to 1/a^2 + 1/b^2 = 2/c^2. This implies 1/a^2, 1/c^2, 1/b^2 are in AP, so a^2, c^2, b^2 are in HP.
Using the perpendicular distance from the center formula, the distance d from the origin to the line bx + ay - ab = 0 is d = |ab| / sqrt(a^2 + b^2). The chord length formula is 2 * sqrt(r^2 - d^2), and setting this equal to sqrt(2)c gives sqrt(c^2 - a^2b^2/(a^2 + b^2)) = sqrt(2)c / 2. Squaring both sides yields c^2 - a^2b^2/(a^2 + b^2) = c^2 / 2, which rearranges to a^2b^2/(a^2 + b^2) = c^2 / 2. This shows that c^2 is the harmonic mean of a^2 and b^2, meaning a^2, c^2, b^2 are in Harmonic Progression.