how to determine the position of points which belong to a circle ?

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jan
jan 2014 年 12 月 22 日
編集済み: jan 2014 年 12 月 23 日
Hi how to determine the points of a circle (their x, y coordinates) if we knew the position of the center and the radius of the circle thanks!!!!

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Roger Stafford
Roger Stafford 2014 年 12 月 22 日
If the center is at (x0,y0) and the radius is r, then
1) if (x-x0)^2 + (y-y0)^2 < r^2, the point (x,y) is inside the circle,
2) if (x-x0)^2 + (y-y0)^2 == r^2, the point (x,y) is on the circle, and
3) if (x-x0)^2 + (y-y0)^2 > r^2, the point (x,y) is outside the circle.
  1 件のコメント
jan
jan 2014 年 12 月 23 日
thanks for this fast response this is what i'm realy looking for just i need now to write the appropriate code

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その他の回答 (1 件)

Image Analyst
Image Analyst 2014 年 12 月 22 日
  6 件のコメント
Image Analyst
Image Analyst 2014 年 12 月 23 日
Well the first way is a totally inefficient way of doing the first method of the FAQ. You can do it just once like the FAQ shows you instead of having a double for loop.
And your second half of your code is just the 4th block of code in the FAQ, just what I recommended to you but you said you didn't use . But you did - you just didn't realize it. Anyway, you don't need to do it both ways.
In your first code it seems like you're trying to get the perimeter pixels, while in the second code you're getting a solid circular disc. I think you should decide whether you want all the pixels in the circle, or just the perimeter pixels, and use the section of the FAQ that does whichever one you want. I would not use your code as-is.
jan
jan 2014 年 12 月 23 日
編集済み: jan 2014 年 12 月 23 日
the second part of my code is just to show you what i want to get as result i.e i used it to shwo you the difference between the both codes. Besides, i had this part from about one year ago and for all these reasons i accepted the first answer and not yours but thanks !!

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