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

# Sphere with given circle as great circle

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2074 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
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the equation of the sphere having the circle x²+y²+z²+7y−2z+2 = 0, 2x+3y+4z−8 = 0 as a great circle.

## Short answer

General sphere through circle: x²+y²+z²+7y-2z+2+λ(2x+3y+4z-8)=0. Centre (-λ, -(7+3λ)/2, (2-4λ)/2) lies on plane 2x+3y+4z-8=0, giving λ=-1. Sphere: x²+y²+z²-2x+4y-6z+10=0.

## Answer

**Given.** Circle: S: x² + y² + z² + 7y − 2z + 2 = 0 with the plane π: 2x + 3y + 4z − 8 = 0.

#### Sphere through the circle
Any sphere through the circle is

x² + y² + z² + 7y − 2z + 2 + λ(2x + 3y + 4z − 8) = 0

```text
x² + y² + z² + 2λx + (7 + 3λ)y + (4λ − 2)z + (2 − 8λ) = 0
```

Its centre is (−λ, −(7 + 3λ)/2, (2 − 4λ)/2).

#### Condition for a great circle
The given circle is a great circle when the centre of the sphere lies on the plane π:

```text
2(−λ) + 3(−(7 + 3λ)/2) + 4((2 − 4λ)/2) − 8 = 0
−2λ − (21 + 9λ)/2 + 4 − 8λ − 8 = 0
```

Multiply by 2: −4λ − 21 − 9λ + 8 − 16λ − 16 = 0, so −29λ − 29 = 0, λ = −1.

#### Equation of the sphere
```text
x² + y² + z² + 7y − 2z + 2 − (2x + 3y + 4z − 8) = 0
x² + y² + z² − 2x + 4y − 6z + 10 = 0
```

Centre (1, −2, 3), radius √(1 + 4 + 9 − 10) = 2. Check: centre on the plane: 2 − 6 + 12 − 8 = 0 ✓.

**Answer: x² + y² + z² − 2x + 4y − 6z + 10 = 0.**

## Check yourself

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

1. How is a sphere through the circle written?
   A. S + lambda(plane) = 0
   B. S x lambda(plane) = 0
   C. S = lambda
   D. S + lambda = 0
   Answer: A. A sphere plus a multiple of the plane contains the circle. (Key point: Family of spheres)

2. When is the circle a great circle?
   A. The sphere's radius equals 1
   B. The centre of the sphere lies on the plane of the circle
   C. The plane passes through the origin
   D. The plane is tangent to the sphere
   Answer: B. A great circle's plane passes through the sphere's centre. (Key point: Great circle condition)

3. What value of lambda results?
   A. 1
   B. -29
   C. -1
   D. 29
   Answer: C. -29 lambda - 29 = 0. (Key point: Lambda)

4. What is the sphere?
   A. x^2 + y^2 + z^2 - 2x + 4y - 6z + 10 = 0
   B. x^2 + y^2 + z^2 + 2x - 4y + 6z + 10 = 0
   C. x^2 + y^2 + z^2 - 2x + 4y - 6z - 10 = 0
   D. x^2 + y^2 + z^2 + 2x + 10y + 2z - 6 = 0
   Answer: A. Substituting lambda = -1 gives centre (1, -2, 3) on the plane, radius 2. (Key point: Sphere)

## Close variants set in other exams

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