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

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

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2078 Kartik, 6 marks
- Video: [Mid-Point Circle Drawing Algorithm - Computer Graphics](https://www.youtube.com/watch?v=jMTSwnsLY1A) — Anisuzzaman Shayak
- 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-10-10
- 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 8 and centre (10, 10).

## Short answer

Use midpoint circle algorithm with p0=-7; including symmetry, first-quadrant points about the origin are (0,8), (1,8), (2,8), (3,7), (4,7), (5,6), (6,5), (7,4), (7,3), (8,2), (8,1), (8,0). Adding center (10,10) gives (10,18), (11,18), (12,18), (13,17), (14,17), (15,16), (16,15), (17,14), (17,13), (18,12), (18,11), (18,10).

## Answer

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

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).

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

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

Adding the centre (10,10):

**Answer (first quadrant): (10,18), (11,18), (12,18), (13,17), (14,17), (15,16), (16,15), (17,14), (17,13), (18,12), (18,11), (18,10).**

## Check yourself

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

1. The decision parameter start for r = 8 is:
   A. p0 = r − 1 = 7
   B. p0 = 1 − r = −7
   C. p0 = 5/4 − r
   D. p0 = 2r
   Answer: B. In the integer midpoint circle form p0 = 1 − r. (Key point: Initial parameter)

2. How are the first-quadrant points obtained from the computed octant?
   A. Each (x,y) gives (−x,y)
   B. Each (x,y) is doubled in x
   C. Each (x,y) also gives (y,x), e.g. (3,7) gives (7,3)
   D. By running the algorithm again from (8,0) with p0 = 1
   Answer: C. Reflecting the octant about y = x completes the first quadrant. (Key point: Getting the first quadrant)

3. After adding the centre (10,10), the point (0,8) becomes:
   A. (10,18)
   B. (10,8)
   C. (18,10)
   D. (8,18)
   Answer: A. (10+0, 10+8) = (10,18). (Key point: Shifting)

## Close variants set in other exams

- [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×

## Related in Computer Graphics

- [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=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=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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