---
title: "Sphere through circle meeting plane in circle of radius 3"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:sphere-through-circle-meeting-plane"
source: https://www.bctengineer.com/q/semester_2/mathematics/sphere-through-circle-meeting-plane
site: BCT Engineering
---

# Sphere through circle meeting plane in circle of radius 3

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2070 Bhadra, 5 marks
- Video: [Obtain equation of the sphere which passes through the circle x^2+y^2=4, z=0 and is cut by the plane](https://www.youtube.com/watch?v=QCBVz59lbcg) — Super Hundred Entrancepedia Institute
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/sphere-through-circle-meeting-plane
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the equation of the sphere through the circle x²+y² = 4, z = 0 and is intersected by the plane x+2y+2z = 0 in a circle of radius 3.

## Short answer

Family x^2+y^2+z^2-4+kz=0 has centre (0,0,-k/2). Distance to plane x+2y+2z=0 is |k|/3. From section radius 3, k^2=36. Spheres: x^2+y^2+z^2±6z-4=0.

## Answer

**Given.** Circle: x² + y² = 4, z = 0. The sphere also meets the plane x + 2y + 2z = 0 in a circle of radius 3.

#### Family of spheres through the circle
The circle lies on the sphere x² + y² + z² = 4 and on the plane z = 0, so every sphere through it is

x² + y² + z² − 4 + kz = 0.

Centre: (0, 0, −k/2). Radius²: R² = 4 + k²/4.

#### Section by the plane x + 2y + 2z = 0
Distance of the centre from the plane:

```text
p = |0 + 0 + 2(−k/2)| / √(1 + 4 + 4) = |k| / 3
```

The circle of section has radius 3, so

```text
R² − p² = 9
4 + k²/4 − k²/9 = 9
(5/36) k² = 5
k² = 36   →   k = 6 or k = −6
```

#### Spheres
- k = 6: **x² + y² + z² + 6z − 4 = 0**
- k = −6: **x² + y² + z² − 6z − 4 = 0**

Check for k = 6: centre (0, 0, −3), R² = 13, p = 2, R² − p² = 9 ✓.

**Answer: x² + y² + z² ± 6z − 4 = 0.**

## Check yourself

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

1. Which family contains all spheres through the circle x^2+y^2 = 4, z = 0?
   A. x^2+y^2+z^2 - 4 + kz = 0
   B. x^2+y^2+z^2 - 4 + k = 0
   C. x^2+y^2 - 4 + kz = 0
   D. x^2+y^2+z^2 + kz = 4k
   Answer: A. S + kP with S the sphere x^2+y^2+z^2 = 4 and P the plane z = 0. (Key point: Family of spheres)

2. For that family, what are the centre and R^2?
   A. (0, 0, k/2), 4 - k^2/4
   B. (0, 0, -k/2), 4 + k^2/4
   C. (0, 0, -k), 4 + k^2
   D. (0, 0, -k/2), 4 - k^2/4
   Answer: B. Completing the square in z gives centre (0, 0, -k/2) and R^2 = 4 + k^2/4. (Key point: Centre and radius)

3. If a plane at distance p from the centre cuts a circle of radius 3, which relation holds?
   A. R^2 + p^2 = 9
   B. R - p = 3
   C. R^2 - p^2 = 9
   D. p^2 - R^2 = 9
   Answer: C. Radius of the section circle squared = R^2 - p^2. (Key point: Section radius relation)

4. Which k values result from 4 + k^2/4 - k^2/9 = 9?
   A. k = 6 or k = -6
   B. k = 6 only
   C. k = 36
   D. k = 3 or k = -3
   Answer: A. (5/36)k^2 = 5 gives k^2 = 36, so both signs are valid. (Key point: Values of k)

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