---
title: "Centre and radius of a circle on sphere"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:centre-radius-circle-sphere"
source: https://www.bctengineer.com/q/semester_2/mathematics/centre-radius-circle-sphere
site: BCT Engineering
---

# Centre and radius of a circle on sphere

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2073 Bhadra, 5 marks; 2071 Magh, 5 marks; 2067 Chaitra, 5 marks (asked 3 times)
- Video: [Class 45 || Find center and radius of circle || Given sphere and plane || BA | BSc 3D Math](https://www.youtube.com/watch?v=h1e2xtmqVI0) — AC Prashant
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/centre-radius-circle-sphere
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Obtain the centre and radius of the circle x²+y²+z²+x+y+z = 4, x+y+z = 0.

## Short answer

The circle lies on sphere x²+y²+z²+x+y+z−4=0 (centre (−1/2,−1/2,−1/2), radius √19/2) and plane x+y+z=0. Distance from centre to plane is √3/2, so circle radius is 2. The foot of perpendicular from centre to plane gives centre (0,0,0).

## Answer

The circle is the intersection of the sphere S: x² + y² + z² + x + y + z − 4 = 0 and the plane x + y + z = 0.

#### Sphere
Compare with x² + y² + z² + 2ux + 2vy + 2wz + d = 0: u = v = w = 1/2, d = −4.

```text
Centre C = (−1/2, −1/2, −1/2)
Radius R = √(u² + v² + w² − d) = √(3/4 + 4) = √19 / 2
```

#### Distance from C to the plane
Plane x + y + z = 0, normal (1, 1, 1):

```text
p = |−1/2 − 1/2 − 1/2| / √3 = (3/2)/√3 = √3/2
```

#### Radius of the circle
```text
r = √(R² − p²) = √(19/4 − 3/4) = √4 = 2
```

#### Centre of the circle
It is the foot of the perpendicular from C to the plane. Points on the perpendicular: (−1/2 + t, −1/2 + t, −1/2 + t). Put in x + y + z = 0:

−3/2 + 3t = 0, t = 1/2. The point is (0, 0, 0).

**Answer: centre (0, 0, 0), radius 2.**

## Check yourself

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

1. For x^2+y^2+z^2+x+y+z = 4, what are the sphere's centre and radius?
   A. Centre (1/2, 1/2, 1/2), radius sqrt(19)/2
   B. Centre (-1/2, -1/2, -1/2), radius sqrt(19)/2
   C. Centre (-1/2, -1/2, -1/2), radius sqrt(13)/2
   D. Centre (-1, -1, -1), radius sqrt(19)
   Answer: B. u = v = w = 1/2, d = -4, so R^2 = 3/4 + 4 = 19/4. (Key point: Sphere data)

2. What is the distance from the sphere centre to x + y + z = 0?
   A. sqrt(3)/2
   B. 3/2
   C. 1/(2 sqrt 3)
   D. sqrt(3)
   Answer: A. |-3/2|/sqrt(3) = sqrt(3)/2. (Key point: Distance to plane)

3. Which formula and value give the circle's radius?
   A. r = sqrt(R^2 + p^2) = sqrt(22)/2
   B. r = R - p = 2
   C. r = sqrt(R^2 - p^2) = 2
   D. r = sqrt(R^2 - p^2) = 4
   Answer: C. r^2 = 19/4 - 3/4 = 4, so r = 2. (Key point: Circle radius)

4. How is the circle's centre found?
   A. As the foot of the perpendicular from the sphere's centre to the plane, here (0, 0, 0)
   B. As the sphere's centre itself
   C. As the point where the plane meets the x-axis
   D. As the midpoint of the intercepts on the axes
   Answer: A. The perpendicular from C meets the plane at t = 1/2, giving (0, 0, 0). (Key point: Circle centre)

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