UPSC NDA I 2019 Mathematics Solved Paper

Concept:The general equation of a sphere is $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$.Its centre is $(-u, -v, -w)$ and its radius is $r = \sqrt{u^2 + v^2 + w^2 - d}$.Explanation:Given equation: $x^{2} + y^{2} + z^{2} - 6x + 8y - 10z + 1 = 0$.Compare it with the general form.The coefficient of $x$ is $-6$, so $2u = -6$ which gives $u = -3$.The coefficient of $y$ is $8$, so $2v = 8$ which gives $v = 4$.The coefficient of $z$ is $-10$, so $2w = -10$ which gives $w = -5$.The constant term is $+1$, so $d = 1$.Now, substitute $u$, $v$, $w$, and $d$ into the radius formula.$r = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{(-3)^2 + (4)^2 + (-5)^2 - 1}$.$= \sqrt{9 + 16 + 25 - 1} = \sqrt{49} = 7$.Answer:$7$ (Option C).