Intersection point between circle and line (Polar coordinates)

Question

I'm wondering if there is a way of finding the intersection point between a line and a circle written in polar coordinates.

% Line
x_line = 10 + r * cos(th);
y_line = 10 + r * sin(th);
%Circle
circle_x = circle_r * cos(alpha);
circle_y = circle_r * sin(alpha);

So far I've tried using the intersect(y_line, circle_y) function without any success. I'm relatively new to MATLAB so bear with me.

Wolfie Wolfie · Accepted Answer · 2017-01-31T12:17:16

I have generalised the below so that other values than a=10 can be used...

a = 10; % constant line offset
th = 0; % constant angle of line 
% rl = ?? - variable to find

% Coordinates of line: 
% [xl, yl] = [ a + rl * cos(th), a + rl * sin(th) ];

rc = 1; % constant radius of circle
% alpha = ?? - variable to find

% Coordinates of circle:
% [xc, yc] = [ rc * cos(alpha), rc * sin(alpha) ];

We want the intersection, so xl = xc, yl = yc

% a + rl * cos(th) = rc * cos(alpha)
% a + rl * sin(th) = rc * sin(alpha)

Square both sides of both equations and sum them. Simplifying sin(a)^2 + cos(a)^2 = 1. Expanding brackets and simplifying further gives

% rl^2 + 2 * a * rl * (cos(th) + sin(th)) + 2 * a - rc^2 = 0

Now you can use the quadratic formula to get the value of rl.

Test discriminant:

dsc = (2 * a * (cos(th) + sin(th)) )^2 - 4 * (2 * a - rc^2);
rl = [];
if dsc < 0
    % no intersection
elseif dsc == 0
    % one intersection at 
    rl = - cos(th) - sin(th);
else
    % two intersection points
    rl = -cos(th) - sin(th) + [ sqrt(dsc)/2, -sqrt(dsc)/2];
end

% Get alpha from an earlier equation
alpha = acos( ( a + rl .* cos(th) ) ./ rc );

Now you have 0, 1 or 2 points of intersection of the line with the circle, from certain known and unknown values about each line. Essentially this is just simultaneous equations, see the start of this article for a basis of the maths https://en.wikipedia.org/wiki/System_of_linear_equations

Intersection point between circle and line (Polar coordinates)

2 Answers