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UPSC 2013 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

A sphere $S$ has points $(0,1,0),(3,-5,2)$ at opposite ends of a diameter. Find the equation of the sphere having the intersection of the sphere $S$ with the plane $5x-2y+4z+7=0$ as a great circle.

Technique

Pencil-of-spheres $S+\lambda\pi=0$; great-circle = “section plane passes through centre”; one linear equation in $\lambda$.

Solution

Strategy. Use the family-of-spheres pencil $S+\lambda\pi=0$ through the intersection circle; pick $\lambda$ so that the new sphere’s centre lies on $\pi$ (the defining property of a great circle).

Step 1 — Equation of $S$ (diameter form)

If $A=(0,1,0)$ and $B=(3,-5,2)$ are diametrically opposite, the sphere $S$ is

which expands to

(Sanity: centre $=\bigl(\tfrac{3}{2},-2,1\bigr)$, radius $=\tfrac{1}{2}|AB|=\tfrac{7}{2}$ since $|AB|^{2}=9+36+4=49$ ✓.)

Step 2 — Family of spheres through $S\cap\pi$

For $\lambda\in\mathbb{R}$, the equation

gives the pencil of spheres containing the circle $S\cap\pi$. Explicitly:

The centre of $(\ast)$ is

Step 3 — Great-circle condition: centre $\in\pi$

A circle on a sphere is a great circle iff its plane passes through the sphere’s centre. So we need $C_\lambda$ to satisfy $\pi$:

Simplify:

Multiply by 2: $15-25\lambda-20\lambda+30=0\;\Rightarrow\;45=45\lambda\;\Rightarrow\;\lambda=1$.

Step 4 — The required sphere

Substitute $\lambda=1$ in $(\ast)$:

Answer