UPSC 2025 Maths Optional Paper 1 Q4a — Step-by-Step Solution

15 marks · Section A

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

Question

Show that there is no tangent plane to the sphere $x^2 + y^2 + z^2 - 4x + 2y - 4z + 4 = 0$ that can be passed through the straight line $\dfrac{x+6}{2} = y + 3 = z + 1$ .

Technique

Take the pencil of planes through the line and impose the plane–sphere tangency condition (distance from centre = radius). A real tangent plane exists iff the resulting quadratic in the pencil parameter has a real root; here the discriminant is negative. As a geometric cross-check, the line passes through the interior of the sphere, so no plane through it can be tangent.

Solution

Step 1 — Sphere data.

$x^2+y^2+z^2-4x+2y-4z+4=0$ has $2u=-4,\,2v=2,\,2w=-4,\,d=4$ , so centre

C=(2,-1,2),\qquad r^2=u^2+v^2+w^2-d=4+1+4-4=5,\qquad r=\sqrt5.

Step 2 — Write the line as the intersection of two planes.

From $\dfrac{x+6}{2}=y+3$ : $\;x+6=2(y+3)\Rightarrow x-2y=0.$ From $y+3=z+1$ : $\;y-z+2=0.$ (Check the point $t=0$ , i.e. $(-6,-3,-1)$ : $x-2y=-6+6=0$ ✓, $y-z+2=-3+1+2=0$ ✓.)

So the line is $\pi_1\equiv x-2y=0,\ \pi_2\equiv y-z+2=0.$

Step 3 — Pencil of planes through the line.

\pi_1+k\,\pi_2=0:\quad x+(k-2)y-kz+2k=0.

Coefficients $(a,b,c)=(1,\,k-2,\,-k)$ , constant $=2k$ .

Step 4 — Impose tangency.

Distance from $C=(2,-1,2)$ to this plane equals $r$ , i.e. $(\,aC_x+bC_y+cC_z+2k\,)^2=r^2(a^2+b^2+c^2)$ :

\big(2+(k-2)(-1)+(-k)(2)+2k\big)^2=5\big(1+(k-2)^2+k^2\big).

Numerator inside: $2-k+2-2k+2k=4-k$ , so $(4-k)^2$ . RHS: $5\big(1+k^2-4k+4+k^2\big)=5(2k^2-4k+5)=10k^2-20k+25.$ Thus

(4-k)^2=10k^2-20k+25\ \Rightarrow\ 16-8k+k^2=10k^2-20k+25\ \Rightarrow\ 9k^2-12k+9=0.

Divide by 3: $3k^2-4k+3=0.$

Step 5 — Discriminant.

\Delta=(-4)^2-4\cdot3\cdot3=16-36=-20<0.

The quadratic has no real root, so no member of the pencil $\pi_1+k\pi_2$ is tangent.

Step 6 — The excluded member $\pi_2$ (the $k\to\infty$ plane).

$\pi_2\equiv y-z+2=0$ has normal $(0,1,-1)$ ; distance from $C$ is $\dfrac{|-1-2+2|}{\sqrt2}=\dfrac{1}{\sqrt2}\approx0.707\neq\sqrt5$ , so $\pi_2$ is not tangent either.

Hence no plane through the given line is tangent to the sphere. $\blacksquare$

Geometric reason (cross-check). Distance from $C$ to the line, with $P_0=(-6,-3,-1)$ and direction $\vec d=(2,1,1)$ :

\frac{|\overrightarrow{P_0C}\times\vec d|}{|\vec d|},\quad \overrightarrow{P_0C}=(8,2,3),\ \overrightarrow{P_0C}\times\vec d=(-1,-2,4),\ \text{dist}=\frac{\sqrt{21}}{\sqrt6}=\sqrt{\tfrac{7}{2}}\approx1.87.

Since $1.87<\sqrt5\approx2.24=r$ , the line pierces the sphere’s interior; every plane containing it cuts the sphere in a circle, never touching it. This corroborates the algebraic result.

Answer

\boxed{\;3k^2-4k+3=0\text{ has discriminant }-20<0,\text{ so no tangent plane to the sphere passes through the line.}\;}