---
title: "Right circular cylinder with guiding circle x²+y²+z²=9"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:right-circular-cylinder-guiding-circle-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/right-circular-cylinder-guiding-circle-2
site: BCT Engineering
---

# Right circular cylinder with guiding circle x²+y²+z²=9

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2068 Magh, 5 marks
- Video: [SOLID GEOMETRY | Cylinder | Equation of Right Circuar Cylinder | Lecture 06 | PRADEEP GIRI SIR](https://www.youtube.com/watch?v=lUepc_KwOBs) — Pradeep Giri Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/right-circular-cylinder-guiding-circle-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Find the equation of a right circular cylinder whose guiding curve is the circle x²+y²+z² = 9, x−y+z = 3.

## Short answer

The cylinder axis passes through the circle centre (1,-1,1) along the plane normal (1,-1,1) and has radius √6. Its equation is x²+y²+z²+xy+yz-zx = 9.

## Answer

A right circular cylinder with a given guiding circle has its axis perpendicular to the plane of the circle and passing through the centre of the circle. Its radius is the radius of the circle.

Circle: x² + y² + z² = 9 (sphere, centre O, radius 3), plane x − y + z = 3.

#### Centre and radius of the circle
The centre is the foot of the perpendicular from O on the plane, along the normal (1, −1, 1): the point (t, −t, t) with t + t + t = 3, so t = 1. Centre C = (1, −1, 1).

Distance from O to the plane = 3/√3 = √3, so

r² = 9 − 3 = 6.

#### Axis
The axis passes through C (and O) with direction ratios (1, −1, 1), the normal of the plane.

#### Equation of the cylinder
Using the distance of P(x, y, z) from the axis through O with direction (1, −1, 1):

```text
(distance)² = OP² − (projection on axis)²
 x² + y² + z² − (x − y + z)²/3 = 6
```
Multiply by 3:

```text
3(x²+y²+z²) − (x² + y² + z² − 2xy + 2xz − 2yz) = 18
2x² + 2y² + 2z² + 2xy + 2yz − 2zx = 18
```

**Answer: x² + y² + z² + xy + yz − zx = 9.**

Check: the point (1+√3, −1+√3, 1) is on the guiding circle (it is on the plane x − y + z = 3 and its squares add to 9). It satisfies x² + y² + z² + xy + yz − zx = 9.

## Check yourself

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

1. The axis of the cylinder with that guiding circle is:
   A. Parallel to the plane
   B. The line joining O to a point on the circle
   C. Through the circle's centre, perpendicular to its plane
   D. The x-axis
   Answer: C. A right circular cylinder's axis is normal to the base circle. (Key point: Axis)

2. The distance from O to x - y + z = 3 is sqrt(3), so r^2 is:
   A. 6
   B. 9
   C. 3
   D. 12
   Answer: A. 9 - 3 = 6. (Key point: Circle radius)

3. A point's squared distance from the axis through O with direction (1, -1, 1) is:
   A. x^2+y^2+z^2 - (x-y+z)^2
   B. x^2+y^2+z^2 - (x-y+z)^2/3
   C. (x-y+z)^2/3
   D. x^2+y^2+z^2 + (x-y+z)^2/3
   Answer: B. OP^2 minus the squared projection on the unit direction. (Key point: Distance from axis)

4. The cylinder is:
   A. x^2+y^2+z^2 + xy + yz - zx = 9
   B. x^2+y^2+z^2 - xy - yz + zx = 9
   C. x^2+y^2+z^2 = 9
   D. x^2+y^2+z^2 + xy + yz - zx = 6
   Answer: A. 3(x^2+y^2+z^2) - (x-y+z)^2 = 18, simplified. (Key point: Equation)

## Close variants set in other exams

- [Right circular cylinder with guiding circle](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aright-circular-cylinder-guiding-circle) — Mathematics, chapter 3, asked 3×

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