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UPSC 2016 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

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

Question

Find the equation of the sphere which passes through the circle $x^2+y^2=4;\ z=0$ and is cut by the plane $x+2y+2z=0$ in a circle of radius 3.

Technique

Pencil of spheres through a circle (sphere $+\lambda\cdot$plane); use $r^2=R^2-d^2$ where $d$ is the centre-to-plane distance, then solve for the parameter.

Solution

Step 1 — Family of spheres through the given circle

The circle $x^2+y^2=4,\ z=0$ is the intersection of the sphere $x^2+y^2+z^2-4=0$ with the plane $z=0$. The family of spheres through it is

Write $\lambda=2f$, so the sphere is $x^2+y^2+z^2+2fz-4=0$ with centre $C=(0,0,-f)$ and radius $R$, where

Step 2 — Perpendicular distance from centre to the cutting plane

Plane $x+2y+2z=0$ has unit-normalizing denominator $\sqrt{1^2+2^2+2^2}=3$. Distance from $C=(0,0,-f)$:

Step 3 — Impose the cut-circle radius $=3$

For a plane cutting a sphere, $r^2=R^2-d^2$ with $r=3$:

Step 4 — Write the spheres

With $2f=\pm6$:

Answer