---
title: "Midpoint circle: r=7, centre (10,15)"
subject: "Computer Graphics"
chapter: 2
key: "semester_5/computer_graphics:circle-r7-center-10-15"
source: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r7-center-10-15
site: BCT Engineering
---

# Midpoint circle: r=7, centre (10,15)

Computer Graphics · Chapter 2: Scan-Conversion

- Asked in: 2081 Chaitra (new course), 5 marks
- Video: [Computer Graphics:Lecture #7: Midpoint circle Example](https://www.youtube.com/watch?v=k5YZBOsw3Jw) — Jyothi Mandala
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/computer_graphics/circle-r7-center-10-15
- Whole chapter: https://www.bctengineer.com/md/semester_5/computer_graphics/chapter-2

## Question

Draw the circle with radius 7 and centre (10,15) using the midpoint (decision parameter) circle algorithm.

## Short answer

Use midpoint circle algorithm with p0=1-r=-6; octant points are (0,7), (1,7), (2,7), (3,6), (4,6), (5,5). Applying eight-way symmetry about (10,15) yields 40 distinct pixels on the circle.

## Answer

**Given:** radius r = 7, centre (10,15). 

Start at (x0, y0) = (0, 7). Initial parameter p0 = 1 − r = -6.

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

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

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 |
|---|---|
| (0,7) | (10,22) (10,22) (10,8) (10,8) (17,15) (3,15) (17,15) (3,15) |
| (1,7) | (11,22) (9,22) (11,8) (9,8) (17,16) (3,16) (17,14) (3,14) |
| (2,7) | (12,22) (8,22) (12,8) (8,8) (17,17) (3,17) (17,13) (3,13) |
| (3,6) | (13,21) (7,21) (13,9) (7,9) (16,18) (4,18) (16,12) (4,12) |
| (4,6) | (14,21) (6,21) (14,9) (6,9) (16,19) (4,19) (16,11) (4,11) |
| (5,5) | (15,20) (5,20) (15,10) (5,10) (15,20) (5,20) (15,10) (5,10) |

**Answer: 40 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 radius 7, p0 = 1 − r equals:
   A. 6
   B. −7
   C. −6
   D. −8
   Answer: C. 1 − 7 = −6. (Key point: Initial parameter)

2. When p < 0, the next point and parameter are:
   A. (x+1, y) and p_new = p + 2x_new + 1
   B. (x+1, y−1) and p_new = p + 2(x_new − y_new) + 1
   C. (x, y−1) and p_new = p + 2y_new
   D. (x+1, y) and p_new = p + 1
   Answer: A. A negative p means the midpoint is inside the circle, so y is kept. (Key point: Update for p < 0)

3. The octant point (0,7) with centre (10,15) is plotted as:
   A. (10,8) only
   B. (10,22) (and symmetric points like (17,15))
   C. (17,22)
   D. (7,15) only
   Answer: B. Add the centre: (10+0, 15+7) = (10,22); swapping gives (10+7, 15+0) = (17,15). (Key point: Shifting to the centre)

## Related in Computer Graphics

- [Midpoint circle algorithm and decision parameter](https://www.bctengineer.com/question.md?key=semester_5%2Fcomputer_graphics%3Amidpoint-circle-derive) — Computer Graphics, chapter 2, asked 8×
- [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×
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- [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

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