UPSC 2024 Maths Optional Paper 1 Q4c — Step-by-Step Solution

15 marks · Section A

Sphere · Analytic Geometry · asked 18× in 14 yrs · Read the full method →

Question

Find the equation of the sphere which touches the plane $3x+2y-z+2=0$ at the point $(1,-2,1)$ and cuts orthogonally the sphere $x^2+y^2+z^2-4x+6y+4=0$ .

Technique

Parametrise the centre on the normal line through the touch-point; express $d$ from the on-sphere condition; close with orthogonality.

Solution

General sphere: $S: x^2+y^2+z^2+2ux+2vy+2wz+d=0$ , centre $C=(-u,-v,-w)$ .

Step 1 — Centre lies on the normal through $P_0=(1,-2,1)$ .

For the sphere to touch the plane at $P_0$ , the radius $\vec{CP_0}$ must be parallel to the plane’s normal $\vec n=(3,2,-1)$ . Write $C=P_0+\mu\vec n$ :

-u=1+3\mu,\quad -v=-2+2\mu,\quad -w=1-\mu.

So $u=-1-3\mu$ , $v=2-2\mu$ , $w=-1+\mu$ .

Step 2 — $P_0$ lies on the sphere.

Condition $1+4+1+2u-4v+2w+d=0$ :

2(-1-3\mu)-4(2-2\mu)+2(-1+\mu)+d=-6\;\Rightarrow\;-12+4\mu+d=-6\;\Rightarrow\;d=6-4\mu.

Step 3 — Orthogonality with $S_2:x^2+y^2+z^2-4x+6y+4=0$ .

For $S_2$ : $u_2=-2,\,v_2=3,\,w_2=0,\,d_2=4$ . Orthogonality condition $2(u u_2+v v_2+w w_2)=d+d_2$ :

2(-2u+3v)=d+4.

Substituting $u,v,d$ :

-4(-1-3\mu)+6(2-2\mu)=(6-4\mu)+4\;\Rightarrow\;16=10-4\mu\;\Rightarrow\;\mu=-\tfrac{3}{2}.

Step 4 — Recover coefficients.

u=-1-3(-\tfrac{3}{2})=\tfrac{7}{2},\quad v=2-2(-\tfrac{3}{2})=5,\quad w=-1+(-\tfrac{3}{2})=-\tfrac{5}{2},\quad d=6-4(-\tfrac{3}{2})=12.

Answer

\boxed{\;x^2+y^2+z^2+7x+10y-5z+12=0.\;}