---
title: "Midpoint circle: r=5, centre (10,5)"
subject: "Computer Graphics"
chapter: 2
key: "semester_5/computer_graphics:circle-r5-center-10-5"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r5-center-10-5
site: BCT Engineering
---

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

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2074 Chaitra, 6 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-r5-center-10-5
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-2

## Question

Calculate the points to draw a circle having radius 5 and center (10, 5) using the midpoint circle algorithm.

## Short answer

Use midpoint circle algorithm with p0=1-r=-4; octant points are (0,5), (1,5), (2,5), (3,4). Applying eight-way symmetry about (10,5) yields 28 distinct pixels on the circle.

## Answer

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

Start at (x0, y0) = (0, 5). Initial parameter p0 = 1 − r = -4.

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

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

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

**Answer: 28 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 the midpoint circle algorithm with r = 5, p0 = 1 − r is:
   A. 4
   B. −4
   C. −5
   D. 6
   Answer: B. 1 − 5 = −4. (Key point: Initial parameter)

2. At (2,5) with p = 4 (≥ 0), the next octant point is:
   A. (3,4)
   B. (3,5)
   C. (2,4)
   D. (3,6)
   Answer: A. p ≥ 0 means x+1 and y−1, so (3,4); p < 0 would keep y. (Key point: Pixel choice)

3. Only one octant is computed; the loop stops when x ≥ y because:
   A. a circle has only two symmetric points
   B. the circle's centre has been reached
   C. the 8-way symmetry supplies the remaining points, with (x,y) giving (±x,±y) and (±y,±x)
   D. p becomes zero
   Answer: C. After x ≥ y the octant is finished; other octants are reflections. (Key point: Symmetry and stopping)

## Close variants set in other exams

- [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×

## Related in Computer Graphics

- [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=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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