UPSC Maths 2025 Paper 1 Q3b — Solution

15 marks · Section A

Question

Find the equations of the spheres which pass through the circle $x^2 + y^2 + z^2 - 2x + 2y + 4z - 3 = 0$ , $2x + y + z = 4$ and touch the plane $3x + 4y = 14$ .

Technique

Write the pencil of spheres through the circle ( $S+\lambda U=0$ ), then impose tangency to the plane (distance from centre = radius) and solve for $\lambda$ .

Solution

Step 1 — Family of spheres through the given circle.

The circle is the intersection of the sphere $S\equiv x^2+y^2+z^2-2x+2y+4z-3=0$ and the plane $U\equiv 2x+y+z-4=0$ . Every sphere through this circle is $S+\lambda U=0:\quad x^2+y^2+z^2+(-2+2\lambda)x+(2+\lambda)y+(4+\lambda)z+(-3-4\lambda)=0.$

Step 2 — Centre and radius.

Centre $C=\left(1-\lambda,\; -\dfrac{2+\lambda}{2},\; -\dfrac{4+\lambda}{2}\right)$ .

Radius $^2 = (1-\lambda)^2+\dfrac{(2+\lambda)^2}{4}+\dfrac{(4+\lambda)^2}{4}-(-3-4\lambda)$ .

Step 3 — Tangency to the plane $3x+4y-14=0$ .

Distance from $C$ to the plane (normal $(3,4,0)$ , $|\,(3,4,0)\,|=5$ ): $\text{dist}=\frac{|3(1-\lambda)+4\!\left(-\frac{2+\lambda}{2}\right)-14|}{5}=\frac{|3-3\lambda-4-2\lambda-14|}{5}=\frac{|-15-5\lambda|}{5}=|3+\lambda|.$

Tangency requires $\text{dist}^2=\text{Radius}^2$ . Substituting and simplifying yields $\lambda^2-2\lambda=0\ \Rightarrow\ \lambda(\lambda-2)=0\ \Rightarrow\ \lambda=0\ \text{ or }\ \lambda=2.$

Step 4 — The two spheres.

$\lambda=0$ : $\;x^2+y^2+z^2-2x+2y+4z-3=0$ (the original sphere $S$ itself), centre $(1,-1,-2)$ , radius $3$ .
$\lambda=2$ : $\;x^2+y^2+z^2+2x+4y+6z-11=0$ , centre $(-1,-2,-3)$ , radius $5$ .

Tangency check. For $\lambda=0$ : dist $=|3+0|=3=$ radius ✓. For $\lambda=2$ : dist $=|3+2|=5=$ radius ✓.

Answer

$\boxed{\;x^2+y^2+z^2-2x+2y+4z-3=0\quad\text{and}\quad x^2+y^2+z^2+2x+4y+6z-11=0.\;}$