UPSC NDA 2 2017 Mathematics Solved Paper

Concept:The center of a sphere passing through the origin and the three axis intercepts is at half the distances of those intercepts from the origin.Explanation:Let the variable plane cut the axes at $A(2x,0,0)$, $B(0,2y,0)$, and $C(0,0,2z)$.The sphere passing through $O(0,0,0)$, $A$, $B$, and $C$ has its center at $(x,y,z)$. This is because the distance from $(x,y,z)$ to each of these four points is the same (all equal the sphere's radius).The equation of the plane in intercept form is $\frac{X}{2x} + \frac{Y}{2y} + \frac{Z}{2z} = 1$.Since it passes through the fixed point $(a,b,c)$, substitute $X=a$, $Y=b$, $Z=c$:$\frac{a}{2x} + \frac{b}{2y} + \frac{c}{2z} = 1$.Multiply both sides by $2$ to get $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$.This relation holds for the coordinates $(x,y,z)$ of the sphere's center. Hence the locus is $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$.Now, we will find point of intersection on X- axis, so $y = 0$, $z = 0$,$\Rightarrow x(x - 2p) = 0$$\text{therefore } x = 0,\ \text{OR}\ x = 2p$Plane cut at point $(2p, 0, 0)$ on X-axisSimilarly, we will find point of intersection on Y- axis, so $x = 0$, $z = 0$,$\text{therefore } y = 0,\ y = 2q$Plane cut at point $(0, 2q, 0)$ on Y-axisAlso, we will find point of intersection on Z- axis, so $x = 0$, $y = 0$,$\text{therefore } z = 0,\ z = 2s$Plane cut at point $(0, 0, 2s)$ on Z-axisAnswer:Option C: $\displaystyle \frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 2$.