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Intersection between a circle and a line.

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Shrishti Yadav
Shrishti Yadav 2021-11-3
编辑： Matt J 2021-11-4
Goal: To get the coordinates of line from every x-y point in the plane intersecting a circle with a given radius.
I am not sure if it is doing that. I was using tan before but I think tan2 works better.I wanted to use equations to solve y= mx and the circle equation but that seemed too complicated. Please let me know if you know of an easier way of doing this.
%calculates the intersection points of a line and a circle
x = -a:b;
y = (-a:b).';
u = numel(x);
Z = zeros(u*u,2);
theta = atan2((y-cy) ,(x-cx)); %cx,cy- center of the circle.
Zx = cx + r*cos(theta);
Zy = cy + r*sin(theta);
Z = [Zx(:) Zy(:)];
  2 个评论
John D'Errico
John D'Errico 2021-11-4
What you are doing is far more complex. Anyway, you CANNOT compute the coordinates of every point that falls inside a region, since there are infinitely many such points. Unless you are asking how to find the points on the line that intersect the perimeter of the circle.
So what is your goal? To get a list of points on the line inside the circle? Or just the two points where the line crosses the perimeter?
Shrishti Yadav
Shrishti Yadav 2021-11-4
To find the coordinates at the intersection of a line from any given (x,y) location to the circle. There are two points for a y = mx intersection with a circle equation.
Thats the goal. In the code I am just using multiple different x,y locations. E.g if x= 1:5 and y =1:5, the total number of unique x,y locations are 25. So for every one of those 25 locations if we draw a line that goes through the center of a circle, it will intersect the circle at two points. Those two points will happen at a coordinates and that is what the code outputs.

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回答（1 个）

Matt J
Matt J 2021-11-4
编辑：Matt J 2021-11-4
One way,
[dx,dy]= ndgrid( (-a:b)-cx , (-a:b)-cy );
factor=r./sqrt(dx.^2+dy.^2);
Z=[dx(:),dy(:)].*factor(:)+[cx,cy];

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