---
title: "Sphere through four points"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:sphere-through-four-points"
source: https://www.bctengineer.com/q/semester_2/mathematics/sphere-through-four-points
site: BCT Engineering
---

# Sphere through four points

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2071 Bhadra, 5 marks
- Video: [Equation of sphere passing through four points](https://www.youtube.com/watch?v=l0tcqylTuNQ) — Suvarna Bhagwat
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/sphere-through-four-points
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the equation of the sphere through the points (0,0,0), (0,1,−1), (−1,2,0) and (1,2,3).

## Short answer

Substitute points into x^2+y^2+z^2+Dx+Ey+Fz+G=0. G=0, then solve to get D=-15/7, E=-25/7, F=-11/7. The sphere equation is 7(x^2+y^2+z^2)-15x-25y-11z=0.

## Answer

**Method.** The general sphere is x² + y² + z² + 2ux + 2vy + 2wz + d = 0, with four unknowns. Four points give four equations. Write D = 2u, E = 2v, F = 2w, G = d:

x² + y² + z² + Dx + Ey + Fz + G = 0.

#### Substitute the points
```text
(0, 0, 0):   G = 0
(0, 1, −1):  0 + 1 + 1 + E − F + G = 0     →  E − F = −2        ... (1)
(−1, 2, 0):  1 + 4 + 0 − D + 2E + G = 0    →  −D + 2E = −5      ... (2)
(1, 2, 3):   1 + 4 + 9 + D + 2E + 3F + G = 0 → D + 2E + 3F = −14 ... (3)
```

#### Solve
Add (2) and (3): 4E + 3F = −19. From (1), F = E + 2, so 4E + 3E + 6 = −19, 7E = −25, E = −25/7. Then F = −25/7 + 2 = −11/7. From (2): D = 2E + 5 = −50/7 + 35/7 = −15/7.

#### Equation
```text
x² + y² + z² − (15/7)x − (25/7)y − (11/7)z = 0
```

Multiplying by 7:

**7(x² + y² + z²) − 15x − 25y − 11z = 0**

Check with (1, 2, 3): 7(14) − 15 − 50 − 33 = 98 − 98 = 0 ✓. With (−1, 2, 0): 35 + 15 − 50 = 0 ✓.

The centre is (15/14, 25/14, 11/14).

## Check yourself

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

1. Why can four non-coplanar points determine a sphere?
   A. The general sphere has four unknowns (u, v, w, d), so four points give four equations
   B. The sphere needs the centre and radius, which are given by any two points
   C. A sphere only needs three points, the fourth is a check
   D. The four points define a plane that contains the sphere
   Answer: A. x^2+y^2+z^2+2ux+2vy+2wz+d = 0 has four constants, fixed by four points. (Key point: Sphere equation form)

2. What does the point (0, 0, 0) lying on the sphere tell you?
   A. D = 0
   B. G (the constant term) = 0
   C. The centre is the origin
   D. E = F
   Answer: B. Substituting the origin leaves only the constant term, so G = 0. (Key point: Use of the origin)

3. After adding (2) and (3) and using F = E + 2, which value of E results?
   A. -19/7
   B. 25/7
   C. -25/7
   D. -5/7
   Answer: C. 4E + 3F = -19 with F = E + 2 gives 7E = -25, so E = -25/7. (Key point: Solving the system)

4. What is the required sphere?
   A. 7(x^2+y^2+z^2) - 15x - 25y - 11z = 0
   B. x^2+y^2+z^2 - 15x - 25y - 11z = 0
   C. 7(x^2+y^2+z^2) + 15x + 25y + 11z = 0
   D. 7(x^2+y^2+z^2) - 15x - 25y - 11z + 7 = 0
   Answer: A. D = -15/7, E = -25/7, F = -11/7, G = 0; multiplying by 7 clears fractions. (Key point: Final equation)

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