---
title: "Midpoint circle: r=10, centre (20,30)"
subject: "Computer Graphics"
chapter: 2
key: "semester_5/computer_graphics:circle-r10-center-20-30"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r10-center-20-30
site: BCT Engineering
---

# Midpoint circle: r=10, centre (20,30)

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2069 Asar, 8 marks
- Video: [Mid point circle algorithm with example in computer graphics | Lec-15](https://www.youtube.com/watch?v=-RX2bygvkZs) — Er Sahil ka Gyan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r10-center-20-30
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-2

## Question

Using the midpoint circle drawing algorithm, find all the points of a circle of radius 10 units in all the quadrants, where the center of the circle is (20, 30).

## Short answer

Use midpoint circle algorithm with p0=1-r=-9; octant points are (0,10), (1,10), (2,10), (3,10), (4,9), (5,9), (6,8), (7,7). Applying eight-way symmetry about (20,30) yields 56 distinct pixels on the circle.

## Answer

**Given:** radius r = 10, centre (20,30). 

Start at (x0, y0) = (0, 10). Initial parameter p0 = 1 − r = -9.

```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 | -9 | (0,10) | (1,10) |
| 1 | -6 | (1,10) | (2,10) |
| 2 | -1 | (2,10) | (3,10) |
| 3 | 6 | (3,10) | (4,9) |
| 4 | -3 | (4,9) | (5,9) |
| 5 | 8 | (5,9) | (6,8) |
| 6 | 5 | (6,8) | (7,7) |
| 7 | 6 | (7,7) | (8,6) |

Octant points about the origin: (0,10), (1,10), (2,10), (3,10), (4,9), (5,9), (6,8), (7,7).

Eight-way symmetry gives, for each octant point (x,y), the points (xc±x, yc±y) and (xc±y, yc±x):

| (x,y) | 8 symmetric points |
|---|---|
| (0,10) | (20,40) (20,40) (20,20) (20,20) (30,30) (10,30) (30,30) (10,30) |
| (1,10) | (21,40) (19,40) (21,20) (19,20) (30,31) (10,31) (30,29) (10,29) |
| (2,10) | (22,40) (18,40) (22,20) (18,20) (30,32) (10,32) (30,28) (10,28) |
| (3,10) | (23,40) (17,40) (23,20) (17,20) (30,33) (10,33) (30,27) (10,27) |
| (4,9) | (24,39) (16,39) (24,21) (16,21) (29,34) (11,34) (29,26) (11,26) |
| (5,9) | (25,39) (15,39) (25,21) (15,21) (29,35) (11,35) (29,25) (11,25) |
| (6,8) | (26,38) (14,38) (26,22) (14,22) (28,36) (12,36) (28,24) (12,24) |
| (7,7) | (27,37) (13,37) (27,23) (13,23) (27,37) (13,37) (27,23) (13,23) |

**Answer: 56 distinct pixels on the whole circle, listed above.**

## Check yourself

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

1. For r = 10 the starting p0 is:
   A. 9
   B. −9
   C. −10
   D. −11
   Answer: B. 1 − 10 = −9. (Key point: Initial parameter)

2. At (3,10) p = 6, the next point (4,9) gives new p of:
   A. 3
   B. 13 (6 + 2·4 + 1 − 2)
   C. −3 (6 + 2(4−9) + 1)
   D. −9
   Answer: C. p_new = p + 2(x_new − y_new) + 1 = 6 − 10 + 1 = −3. (Key point: Update for p >= 0)

3. Why does the octant end at (7,7)?
   A. p becomes negative
   B. y reaches zero
   C. x equals the radius
   D. Once x ≥ y, the remaining points are mirror images given by symmetry
   Answer: D. The loop runs while x < y; (7,7) is the point on the line y = x. (Key point: Octant end)

## Related in Computer Graphics

- [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=6, centre (20,10)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r6-center-20-10) — 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=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×
- [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=7, centre (10,15)](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Acircle-r7-center-10-15) — Computer Graphics, chapter 2

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