UPSC NDA I 2016 Mathematics Solved Paper

Concept:The intercepts cut by a circle on the axes are the lengths of the chords when $y=0$ (x‑intercept) and $x=0$ (y‑intercept).Explanation:From the earlier problem, the circle’s equation has $c=0$, $g=\frac{a^{2}+b^{2}}{2(b-a)}$, and $f=\frac{a^{2}+b^{2}}{2(a-b)}$.For the x‑intercept, put $y=0$: $x^{2}+2gx=0$. This gives roots $x=0$ and $x=-2g=\frac{a^{2}+b^{2}}{a-b}$.For the y‑intercept, put $x=0$: $y^{2}+2fy=0$. This gives roots $y=0$ and $y=-2f=\frac{a^{2}+b^{2}}{b-a}$.The square of the x‑intercept is $\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$ and the square of the y‑intercept is $\left(\frac{a^{2}+b^{2}}{b-a}\right)^{2}$.Since $(b-a)^{2}=(a-b)^{2}$, both squares are equal.Therefore, the sum of the squares is $2\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$.Answer:B. $2\left(\frac{a^{2}+b^{2}}{a-b}\right)^{2}$