UPSC NDA Math Model Paper 6

Given: Sphere passes through $(0,0,0),(2,1,-1),(1,5,-4)$ and $(-2,4,-6)$ General equation of sphere, $x^{2}+y^{2}+z^{2}+2 u x+2 v y$ $+2 w z+d=0$ .......(1) Put $(0,0,0)$ in the equation (1) we get, $0+0+0+0+0+0+d=0$ $\Rightarrow d=0$ Now, equation becomes, $x^{2}+y^{2}+z^{2}+2 u x+2 v y$ $+2 w z=0$ ......(2) Put $(2,1,-1)$ in equation ( 2 ) we get, $2^{2}+1^{2}+(-1)^{2}+2 u(2)+2 v(1)$ $+2 w(-1)=0$ $\Rightarrow 6 + 4u + 2v - 2w = 0$ .......(3) Put (1, 5, -4) in equation (2) $1^{2}+5^{2}+(-4)^{2}+2 u(1)+2 v(5)$ $+2 w(-4)=0$ $\Rightarrow 42 + 2u + 10v - 8w = 0$ .......(4) Put (-2, 4, -6) in equation (2) $(-2)^{2}+4^{2}+(-6)^{2}+2 u(-2)+2 v(4)$ $+2 w(-6)=0$ ⇒ 56 - 4u + 8v - 12w = 0 ......(5) Now, adding equation (3) and (5) 62 + 10v - 14w = 0 .......(6) Divide equation (5) and add it to equation (4) 28 - 2u + 4v - 6w + 42 + 2u + 10v - 8w = 0 ⇒ 70 + 14v - 14w = 0 .......(7) Now, subtracting (6) from (7) 8 + 4v = 0 ⇒ v = -2 Put v = -2 in equation (7) 70 + 14(-2) - 14w = 0 ⇒ 14w = 42 ⇒ w = 3 Put v = -2 and w = 3 in equation (3) 6 + 4u + 2(-2) - 2(3) = 0 ⇒ 4u = 4 ⇒ u = 1 Equation of circle is given as, $x^{2}+y^{2}+z^{2}+2 x-4 y+6 z=0$ Radius of circle, $r = \sqrt{u^{2}+v^{2}+z^{2}-d}$ $\Rightarrow r = \sqrt{1^{2}+(-2)^{2}+3^{2}}$ $\Rightarrow r = \sqrt{14}$