---
title: "Sphere with given circle as great circle"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:sphere-given-circle-as-great-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/sphere-given-circle-as-great-2
site: BCT Engineering
---

# Sphere with given circle as great circle

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2069 Bhadra, 5 marks
- Video: [|| To find the equation of a sphere for which the given circle is a great circle || ](https://www.youtube.com/watch?v=PyWJ-0TC2aM) — dy by dx
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/sphere-given-circle-as-great-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Obtain the equation of the sphere through the circle x²+y²+z² = 9, x−2y+2z = 5 as a great circle.

## Short answer

Using the family S + kP = 0 with centre (-k/2, k, -k), imposing that the centre lies on the plane yields k = -10/9. Substituting gives the sphere 9(x^2+y^2+z^2) - 10x + 20y - 20z - 31 = 0.

## Answer

A sphere has a given circle as a great circle when the circle lies on the sphere and the centre of the sphere lies in the plane of the circle.

The circle is x² + y² + z² = 9, x − 2y + 2z = 5.

#### Equation of the family
Every sphere through the circle is

S + kP = 0, i.e. x² + y² + z² − 9 + k(x − 2y + 2z − 5) = 0.

Rewrite it as

x² + y² + z² + kx − 2ky + 2kz − 9 − 5k = 0.

Its centre is (−k/2, k, −k).

#### Centre on the plane
The circle is a great circle, so the centre must lie on the plane x − 2y + 2z = 5:

```text
(−k/2) − 2(k) + 2(−k) = 5
−9k/2 = 5   →   k = −10/9
```

#### Required sphere
```text
x² + y² + z² − 9 − (10/9)(x − 2y + 2z − 5) = 0
9(x² + y² + z²) − 81 − 10x + 20y − 20z + 50 = 0
```

**Answer: 9(x² + y² + z²) − 10x + 20y − 20z − 31 = 0.**

Check: the centre is (5/9, −10/9, 10/9), and 5/9 + 20/9 + 20/9 = 5, so it lies on the plane.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. A circle is a great circle of a sphere when:
   A. The sphere's centre lies in the plane of the circle
   B. The sphere's centre lies on the circle
   C. The plane passes through the origin
   D. The sphere has radius equal to the circle's
   Answer: A. The centre in the plane means the plane cuts the sphere through its centre. (Key point: Great circle condition)

2. Which family contains spheres through the given circle?
   A. S + kP^2 = 0
   B. S + kP = 0, with S the sphere and P the plane
   C. SP = k
   D. S = kP with no constant
   Answer: B. S + kP = 0 passes through every point where S = 0 and P = 0. (Key point: Family)

3. The member x^2+y^2+z^2 + kx - 2ky + 2kz - 9 - 5k = 0 has centre:
   A. (k/2, -k, k)
   B. (-k, 2k, -2k)
   C. (-k/2, k, -k)
   D. (-k/2, k, 2k)
   Answer: C. Centre is (-coefficients/2) = (-k/2, k, -k). (Key point: Centre of the family member)

4. Putting the centre in x - 2y + 2z = 5 gives:
   A. k = -10/9
   B. k = 10/9
   C. k = -2/9
   D. k = -5/9
   Answer: A. -k/2 - 2k - 2k = -9k/2 = 5, so k = -10/9. (Key point: Value of k)

## Close variants set in other exams

- [Sphere with circle as great circle](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Asphere-circle-as-great-circle) — Mathematics, chapter 3, asked 3×

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