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UPSC 2022 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

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

Question

A variable plane passes through a fixed point $(a,b,c)$ and meets the axes at points $A,B,C$ respectively. Find the locus of the centre of the sphere passing through the points $O,A,B,C$, where $O$ is the origin.

Technique

Intercept form of plane; sphere through O has zero constant term; coefficients determined by passing through A, B, C; centre at $(\alpha/2,\beta/2,\gamma/2)$; eliminate $\alpha,\beta,\gamma$ using the plane-through-$(a,b,c)$ constraint.

Solution

Setup. Let the plane be $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=1$ (intercept form). It meets axes at $A=(\alpha,0,0)$, $B=(0,\beta,0)$, $C=(0,0,\gamma)$.

The plane passes through $(a,b,c)$, so

Step 1 — Sphere through $O,A,B,C$

A sphere through $O=(0,0,0)$ has equation

(constant term is zero since it passes through the origin).

For the sphere to pass through $A=(\alpha,0,0)$: $\alpha^2+2u\alpha=0\Rightarrow u=-\alpha/2$.

Similarly $v=-\beta/2$, $w=-\gamma/2$.

So the sphere is

Step 2 — Centre of the sphere

The general equation $x^2+y^2+z^2+2ux+2vy+2wz=0$ has centre $(-u,-v,-w)$. Here:

Let $(X,Y,Z)$ be the centre: $X=\alpha/2,\;Y=\beta/2,\;Z=\gamma/2$, i.e. $\alpha=2X,\;\beta=2Y,\;\gamma=2Z$.

Step 3 — Eliminate $\alpha,\beta,\gamma$ using $(\star)$

Substitute $\alpha=2X,\beta=2Y,\gamma=2Z$ into $\dfrac{a}{\alpha}+\dfrac{b}{\beta}+\dfrac{c}{\gamma}=1$:

i.e.

Replace $(X,Y,Z)$ with general $(x,y,z)$ (the locus equation):

Answer