← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

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

Question

For what positive value of $a$, the plane $ax-2y+z+12=0$ touches the sphere $x^2+y^2+z^2-2x-4y+2z-3=0$, and hence find the point of contact.

Technique

Standard sphere–plane tangency: complete squares for centre/radius, set distance-from-centre equal to radius, solve quadratic in $a$; foot of perpendicular for contact point.

Solution

Step 1 — Identify centre and radius of the sphere

Complete squares:

So centre $C=(1,2,-1)$, radius $r=3$.

Step 2 — Tangency condition: distance from $C$ to plane equals $r$

The plane is $ax-2y+z+12=0$. Distance:

Set $d=3$:

Quadratic: $a=\dfrac{7\pm\sqrt{49+32}}{8}=\dfrac{7\pm 9}{8}$. So $a=2$ or $a=-1/4$.

Positive value: $\boxed{a=2}$.

Step 3 — Find the point of contact

The point of contact is the foot of perpendicular from $C=(1,2,-1)$ to the plane $2x-2y+z+12=0$.

Normal direction: $\vec n=(2,-2,1)$, $|\vec n|=3$.

Signed distance from $C$ to plane: $\dfrac{2(1)-2(2)+1(-1)+12}{3}=\dfrac{2-4-1+12}{3}=\dfrac{9}{3}=3$.

Move from $C$ along $-\vec n/|\vec n|$ by distance 3 (negative because the signed numerator is positive, meaning $C$ is on the $+\vec n$ side, and the foot is in the $-\vec n$ direction):

Step 4 — Verify

• $P$ on plane: $2(-1)-2(4)+(-2)+12=-2-8-2+12=0$ ✓.
• $|CP|^2=(-2)^2+2^2+(-1)^2=4+4+1=9$, $|CP|=3=r$ ✓.

Answer