← 2014 Paper 1

UPSC 2014 Maths Optional Paper 1 Q4a-i — Step-by-Step Solution

10 marks · Section A

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

Question

Find the co-ordinates of the points on the sphere $x^{2}+y^{2}+z^{2}-4x+2y=4$, the tangent planes at which are parallel to the plane $2x-y+2z=1$.

Technique

Sphere → standard form via completing the square; radius vector parallel to given plane’s normal.

Solution

Strategy. Identify sphere centre and radius. The tangent plane at any point is perpendicular to the radius vector at that point. For the tangent plane to be parallel to the given plane, the radius vector must be parallel to the given plane’s normal.

Step 1 — Sphere in standard form

Centre $C=(2,-1,0)$, radius $R=3$.

Step 2 — Direction of given plane’s normal

Plane $2x-y+2z=1$ has normal $\vec n=(2,-1,2)$. Magnitude $|\vec n|=\sqrt{4+1+4}=3$.

Step 3 — Find points

The radius vector from $C$ to a point $P=(x_0,y_0,z_0)$ on the sphere is $\vec{CP}=(x_0-2,y_0+1,z_0)$. For the tangent plane at $P$ to be parallel to $2x-y+2z=1$, $\vec{CP}$ must be parallel to $\vec n$:

Magnitude: $|\vec{CP}|=R=3$, $|t\vec n|=3|t|$. So $3|t|=3$, $t=\pm 1$.

Case $t=+1$: $P=C+\vec n=(2+2,-1-1,0+2)=(4,-2,2)$.

Case $t=-1$: $P=C-\vec n=(2-2,-1+1,0-2)=(0,0,-2)$.

Answer