UPSC 2018 Maths Optional Paper 1 Q3d — Step-by-Step Solution

12 marks · Section A

Sphere · Analytic Geometry · asked 17× in 13 yrs · Read the full method →

Question

Find the equation of the sphere in $xyz$ -plane passing through the points $(0,0,0),\ (0,1,-1),\ (-1,2,0)$ and $(1,2,3)$ .

General-equation method $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ ; passage through the origin gives $d=0$ , then solve a $3\times3$ linear system.

A sphere has the form

x^2+y^2+z^2+2ux+2vy+2wz+d=0.

Four points determine $u,v,w,d$ .

$(0,0,0)$ : $0+0+0+0+0+0+d=0\Rightarrow d=0.$ So the sphere passes through $O$ and the equation simplifies to $x^2+y^2+z^2+2ux+2vy+2wz=0$ .

$(0,1,-1)$ : $0+1+1+2u(0)+2v(1)+2w(-1)=0\Rightarrow 2+2v-2w=0\Rightarrow v-w=-1.$

$(-1,2,0)$ : $1+4+0+2u(-1)+2v(2)+0=0\Rightarrow 5-2u+4v=0\Rightarrow -2u+4v=-5.$

$(1,2,3)$ : $1+4+9+2u(1)+2v(2)+2w(3)=0\Rightarrow 14+2u+4v+6w=0\Rightarrow u+2v+3w=-7.$

From the first relation $w=v+1$ . Substitute into the third: $u+2v+3(v+1)=-7\Rightarrow u+5v=-10$ . With $-2u+4v=-5$ (i.e. $u=2v+\tfrac52$ ):

\big(2v+\tfrac52\big)+5v=-10\Rightarrow 7v=-\tfrac{25}{2}\Rightarrow v=-\tfrac{25}{14}.

Then $u=2(-\tfrac{25}{14})+\tfrac52=-\tfrac{50}{14}+\tfrac{35}{14}=-\tfrac{15}{14}$ , and $w=v+1=-\tfrac{25}{14}+\tfrac{14}{14}=-\tfrac{11}{14}.$

With $2u=-\tfrac{15}{7},\ 2v=-\tfrac{25}{7},\ 2w=-\tfrac{11}{7}$ :

x^2+y^2+z^2-\tfrac{15}{7}x-\tfrac{25}{7}y-\tfrac{11}{7}z=0,

or clearing denominators,

\boxed{\;7\big(x^2+y^2+z^2\big)-15x-25y-11z=0.\;}