---
title: "Midpoint circle: r=10, centre (50,50)"
subject: "Computer Graphics"
chapter: 2
key: "semester_5/computer_graphics:circle-r10-center-50-50-octant"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r10-center-50-50-octant
site: BCT Engineering
---

# Midpoint circle: r=10, centre (50,50)

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2066 Magh (old course)
- Video: [mid point algorithm computer graphics | midpoint circle algorithm | computer graphics | hindi](https://www.youtube.com/watch?v=JoBLO25izd8) — computer science
- 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-50-50-octant
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-2

## Question

Calculate all the pixels of a circle with radius = 10 and center at (50, 50) in the first octant starting from (50, 60) proceeding to positive x axis direction.

## Short answer

Use midpoint circle algorithm from (0,10) with p0=-9; first-octant points are (0,10), (1,10), (2,10), (3,10), (4,9), (5,9), (6,8), (7,7). Adding center (50,50) gives (50,60), (51,60), (52,60), (53,60), (54,59), (55,59), (56,58), (57,57).

## Answer

**Given:** radius r = 10, centre (50,50). The first octant starts at (50,60), the top of the circle, and proceeds in the +x direction.

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).

Shift by (50,50) (add xc to x and yc to y):

| k | (x,y) in octant | plotted (x+xc, y+yc) |
|---|---|---|
| 0 | (0,10) | (50,60) |
| 1 | (1,10) | (51,60) |
| 2 | (2,10) | (52,60) |
| 3 | (3,10) | (53,60) |
| 4 | (4,9) | (54,59) |
| 5 | (5,9) | (55,59) |
| 6 | (6,8) | (56,58) |
| 7 | (7,7) | (57,57) |

**Answer: first-octant pixels: (50,60), (51,60), (52,60), (53,60), (54,59), (55,59), (56,58), (57,57).**

## Check yourself

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

1. With centre (50,50) and r = 10, the first-octant start (0,10) is plotted as:
   A. (60,50)
   B. (50,60)
   C. (50,10)
   D. (10,50)
   Answer: B. Add the centre: (50+0, 50+10). (Key point: Start point)

2. The starting decision parameter is:
   A. p0 = −9
   B. p0 = 9
   C. p0 = −10
   D. p0 = 1
   Answer: A. p0 = 1 − r = −9. (Key point: Initial parameter)

3. At (3,10), p = 6 ≥ 0. What is the next plotted pixel?
   A. (54,60)
   B. (53,59)
   C. (54,61)
   D. (54,59)
   Answer: D. p ≥ 0 gives (4,9), shifted by (50,50) to (54,59). (Key point: When y drops)

## Close variants set in other exams

- [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×

## 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=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

## Related in other subjects

- [Centre and radius of a circle on sphere](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acentre-radius-circle-sphere) — Mathematics, chapter 3, asked 3×
- [Centre and radius of circle of sphere section](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acentre-radius-circle-sphere-section) — Mathematics, chapter 3, asked 3×
- [Dirichlet integral over positive octant region](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Adirichlet-integral-over-positive-octant) — Mathematics, chapter 2, asked 3×