MATLAB: How to check if a line segment intersects a circle

circlegeometryImage Processing Toolboxintersection of line and circleline

is there any formula to check whether a line intersects a circle in matlab.can someone help me out. suppose i have got a line segemtn with ends x,y and x1,y2 and there is a circle with centre h,k with radius r. how can i know if the line segement intersects the circle or not

Best Answer

Calculate the distance between the line and the center of the circle:

P1 = [-2, -2];  % Point 1 of the line
P2 = [2, 2];    % Point 2 of the line
C  = [0, 0];    % Center of circle
R  = 2;         % Radius of circle
P12 = P2 - P1;
N   = P12 / norm(P12);  % Normalized vector in direction of the line
P1C = C - P1;           % Line from one point to center
v   = abs(N(1) * P1C(2) - N(2) * P1C(1));  % Norm of the 2D cross-product
doIntersect = (v <= R);

The geometrical explanation is easy: The cross-product replies the area of the parallelogram build by the normal of the line and the vector fro P1 to C. The area of this parallelogram is identical to the area of the rectangle build by the normal N and the vector orthogonal to N through C. The value of this area is the distance multiplied by 1 (the length of the N). Summary: The distance between the line and a point in 2D is the absolute value of the 3rd component of the cross-product between N and the vector from P1 (or P2) to C.

All you have to do is to compare, if this distance is smaller or equal to the radius.

In 3D the norm of the 3D cross-product is needed:

v = norm(cross(N, P1C));

Related Solutions

MATLAB: For intersecting circles, how to remove overlap and have chord visible? Also, how to randomly generate circle position in design space

How are you drawing your circles?

However you are trying to draw it you'll probably end up having x and y points representing the 2 circles.

xCenter1 = 7;
yCenter1 = 7;
xCenter2 = 12;
yCenter2 = 10;
theta = 0 : 0.01 : 2*pi;
radius1 = 5;
radius2 = 6;
%generate 2 circles with parameters above.
  x1 = radius1 * cos(theta) + xCenter;
  y1 = radius1 * sin(theta) + yCenter;
x2 = radius2 * cos(theta) + xCenter2;
y2 = radius2 * sin(theta) + yCenter2;
%%find distance from each circle center to the other circle's infringing points.
dC1 = sqrt((x2-xCenter1).^2+(y2-yCenter1).^2)>=radius1;
dC2 = sqrt((x1-xCenter2).^2+(y1-yCenter2).^2)>=radius2;
plot(x1(dC2), y1(dC2),'b.',x2(dC1),y2(dC1),'r.');
axis square;
xlim([0 20]);
ylim([0 20]);
grid on;

Looking at your sample picture it is easy to see that the infringing portion of circle 2 into circle 1 are points that are within the radius of circle 1. (vice versa for circle 2) With this we can calculate the distance of all points of circle 2 to the center of circle 1 and compare it to the radius.

MATLAB: Distance between two points on the sphere.

I guess you want to find the shortest distance along the surface of the sphere, not just the euclidean distance between the points. The shortest path between two points on a sphere is always located on a great circle, which is thus a "great arc". You can find Roger Staffords mathematically robust method here:

https://www.mathworks.com/matlabcentral/newsreader/view_thread/278872

For convenience I repeat it here too:

" P1 = [x1,y1,z1] and P2 = [x2,y2,z2] are two vectors pointing from the center of the sphere to the two given points (x1,y1,z1) and (x2,y2,z2) on the sphere, what is the shortest great circle distance d between them?"

 d = radius * atan2(norm(cross(P1,P2)),dot(P1,P2));

If you want an example of how to use this to generate points along a great arc, then see the Mfile colornames_cube in my FEX submission colornames. The nested function cncDemoClBk at the end of the file steps along a great arc, rotating the axes as it goes.

Best Answer

Related Solutions

MATLAB: For intersecting circles, how to remove overlap and have chord visible? Also, how to randomly generate circle position in design space

MATLAB: Distance between two points on the sphere.

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