---
title: "Midpoint circle: r=8, centre (-5,10)"
subject: "Computer Graphics"
chapter: 2
key: "semester_5/computer_graphics:circle-r8-center-neg5-10"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r8-center-neg5-10
site: BCT Engineering
---

# Midpoint circle: r=8, centre (-5,10)

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2072 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r8-center-neg5-10
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-2

## Question

Calculate the points in the circumference of the circle having radius 8 units and center at (-5, 10) using the midpoint circle algorithm.

## Short answer

Use midpoint circle algorithm with p0=1-r=-7; octant points are (0,8), (1,8), (2,8), (3,7), (4,7), (5,6). Applying eight-way symmetry about (-5,10) yields 44 distinct pixels on the circle.

## Answer

**Given:** radius r = 8, centre (-5,10). 

Start at (x0, y0) = (0, 8). Initial parameter p0 = 1 − r = -7.

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

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

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 |
|---|---|
| (0,8) | (-5,18) (-5,18) (-5,2) (-5,2) (3,10) (-13,10) (3,10) (-13,10) |
| (1,8) | (-4,18) (-6,18) (-4,2) (-6,2) (3,11) (-13,11) (3,9) (-13,9) |
| (2,8) | (-3,18) (-7,18) (-3,2) (-7,2) (3,12) (-13,12) (3,8) (-13,8) |
| (3,7) | (-2,17) (-8,17) (-2,3) (-8,3) (2,13) (-12,13) (2,7) (-12,7) |
| (4,7) | (-1,17) (-9,17) (-1,3) (-9,3) (2,14) (-12,14) (2,6) (-12,6) |
| (5,6) | (0,16) (-10,16) (0,4) (-10,4) (1,15) (-11,15) (1,5) (-11,5) |

**Answer: 44 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 = 8, p0 =
   A. 7
   B. −8
   C. −9
   D. −7
   Answer: D. 1 − 8 = −7. (Key point: Initial parameter)

2. At (2,8) with p = 1, the next point is:
   A. (3,7), with p = 1 + 2(3−7) + 1 = −6
   B. (3,8), with p = 6
   C. (3,7), with p = 1 + 2(3) + 1 = 8
   D. (2,7), with p = −6
   Answer: A. p ≥ 0 gives (x+1, y−1) and p_new = p + 2(x_new − y_new) + 1. (Key point: Pixel choice)

3. With centre (−5,10), the octant point (0,8) gives:
   A. (5,18)
   B. (−5,18)
   C. (−5,2) only
   D. (0,18)
   Answer: B. Add the centre to (x,y): (−5+0, 10+8) = (−5,18); the centre may be negative. (Key point: Centre shift)

## Close variants set in other exams

- [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×

## 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=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=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=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×
- [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×

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