---
title: "Midpoint circle: r=6, centre (20,10)"
subject: "Computer Graphics"
chapter: 2
key: "semester_5/computer_graphics:circle-r6-center-20-10"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r6-center-20-10
site: BCT Engineering
---

# Midpoint circle: r=6, centre (20,10)

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2070 Chaitra, 4 marks
- Video: [Mid Point Circle Drawing Algorithm Numerical 1 Explained in Hindi l Computer Graphics Series](https://www.youtube.com/watch?v=00zhJvHuLcY) — 5 Minutes Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r6-center-20-10
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-2

## Question

Using the midpoint circle drawing algorithm, calculate the coordinates on the first quadrant of a circle having radius 6 and centre (20, 10).

## Short answer

Using midpoint circle algorithm with r=6 and center (20,10), the first-quadrant points are (20,16), (21,16), (22,16), (23,15), (24,14), (25,13), (26,12), (26,11), (26,10).

## Answer

**Given:** radius r = 6, centre (20,10).

Start at (x0, y0) = (0, 6). Initial parameter p0 = 1 − r = −5.

```text
p < 0 : next point (x+1, y),     p_new = p + 2x_new + 1
p ≥ 0 : next point (x+1, y−1),   p_new = p + 2(x_new − y_new) + 1
Stop when x ≥ y
```

Decision parameter table (circle centred at the origin; p_k decides the next point):

| k | p_k | (x_k, y_k) | next (x, y) |
|---|---|---|---|
| 0 | -5 | (0,6) | (1,6) |
| 1 | -2 | (1,6) | (2,6) |
| 2 | 3 | (2,6) | (3,5) |
| 3 | 0 | (3,5) | (4,4) |
| 4 | 1 | (4,4) | (5,3) |

Octant points about the origin: (0,6), (1,6), (2,6), (3,5), (4,4).

By symmetry, each point (x,y) also gives (y,x), so the first-quadrant points about the origin, in order along the arc, are:

(0,6), (1,6), (2,6), (3,5), (4,4), (5,3), (6,2), (6,1), (6,0)

Adding the centre (20,10):

**Answer (first quadrant): (20,16), (21,16), (22,16), (23,15), (24,14), (25,13), (26,12), (26,11), (26,10).**

## Check yourself

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

1. For r = 6, p0 =
   A. −5
   B. 5
   C. −6
   D. −7
   Answer: A. 1 − 6 = −5. (Key point: Initial parameter)

2. At (3,5), p = 0. Which point is next?
   A. (4,5), since p is not negative
   B. (4,4), since p ≥ 0 means x+1 and y−1
   C. (3,4)
   D. (4,6)
   Answer: B. The rule is p < 0 keep y, otherwise decrement y. (Key point: p = 0 case)

3. How are the points like (6,0), (6,1), (6,2) obtained?
   A. By running the loop with x decreasing
   B. By adding the radius to x
   C. By reflecting octant points (x,y) to (y,x), e.g. (2,6) gives (6,2)
   D. By negating the y values
   Answer: C. Reflecting about y = x gives the second octant of the quadrant. (Key point: First quadrant)

## Close variants set in other exams

- [Midpoint circle: r=8, centre (10,10)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r8-center-10-10) — Computer Graphics, chapter 2, asked 2×

## Related in Computer Graphics

- [Midpoint circle: r=10, centre (20,30)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r10-center-20-30) — Computer Graphics, chapter 2
- [Midpoint circle: r=30, centre (10,20)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r30-center-10-20-octant) — Computer Graphics, chapter 2, asked 2×
- [Midpoint circle: r=5, centre (10,5)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r5-center-10-5) — Computer Graphics, chapter 2, asked 2×
- [Midpoint circle: r=8, centre (-5,10)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r8-center-neg5-10) — Computer Graphics, chapter 2, asked 2×
- [Midpoint circle: r=7, centre (10,15)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r7-center-10-15) — Computer Graphics, chapter 2
- [Midpoint circle: r=10, centre (50,50)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r10-center-50-50-octant) — Computer Graphics, chapter 2, asked 2×

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