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How do you calculate the intersection point of a line and a circle given a line starting point Lx Ly and an angle from north a the circle location Cx Cy and its radius r?
Wiki User
∙ 10y ago
The angle of the line is a degrees from North so the angle from East is (90 - a) degrees. Therefore gradient of the line is tan(90-a) = 1/tan(a).
Then any point on the line has coordinates (x, y) = (Lx + ksina, Ly + kcosa).
The coordinates of any point on the circle satisfy (x - Cx)2 + (y - Cy)2 = r2
The points of intersection must satisfy both equations so that:
(Lx + ksina - Cx)2 + (Ly + kcosa - Cy)2 = r2
This equation needs to be solved for k - everything else has a known numeric value. Since it is a quadratic, in k, there will be 0, 1 or 2 solutions for k depending on the line not intersecting the circle at all, being tangent to it and going through the circle.
The angle of the line is a degrees from North so the angle from East is (90 - a) degrees. Therefore gradient of the line is tan(90-a) = 1/tan(a).
Then any point on the line has coordinates (x, y) = (Lx + ksina, Ly + kcosa).
The coordinates of any point on the circle satisfy (x - Cx)2 + (y - Cy)2 = r2
The points of intersection must satisfy both equations so that:
(Lx + ksina - Cx)2 + (Ly + kcosa - Cy)2 = r2
This equation needs to be solved for k - everything else has a known numeric value. Since it is a quadratic, in k, there will be 0, 1 or 2 solutions for k depending on the line not intersecting the circle at all, being tangent to it and going through the circle.
The angle of the line is a degrees from North so the angle from East is (90 - a) degrees. Therefore gradient of the line is tan(90-a) = 1/tan(a).
Then any point on the line has coordinates (x, y) = (Lx + ksina, Ly + kcosa).
The coordinates of any point on the circle satisfy (x - Cx)2 + (y - Cy)2 = r2
The points of intersection must satisfy both equations so that:
(Lx + ksina - Cx)2 + (Ly + kcosa - Cy)2 = r2
This equation needs to be solved for k - everything else has a known numeric value. Since it is a quadratic, in k, there will be 0, 1 or 2 solutions for k depending on the line not intersecting the circle at all, being tangent to it and going through the circle.
The angle of the line is a degrees from North so the angle from East is (90 - a) degrees. Therefore gradient of the line is tan(90-a) = 1/tan(a).
Then any point on the line has coordinates (x, y) = (Lx + ksina, Ly + kcosa).
The coordinates of any point on the circle satisfy (x - Cx)2 + (y - Cy)2 = r2
The points of intersection must satisfy both equations so that:
(Lx + ksina - Cx)2 + (Ly + kcosa - Cy)2 = r2
This equation needs to be solved for k - everything else has a known numeric value. Since it is a quadratic, in k, there will be 0, 1 or 2 solutions for k depending on the line not intersecting the circle at all, being tangent to it and going through the circle.
Wiki User
∙ 10y ago
Wiki User
∙ 10y agoThe angle of the line is a degrees from North so the angle from East is (90 - a) degrees. Therefore gradient of the line is tan(90-a) = 1/tan(a).
Then any point on the line has coordinates (x, y) = (Lx + ksina, Ly + kcosa).
The coordinates of any point on the circle satisfy (x - Cx)2 + (y - Cy)2 = r2
The points of intersection must satisfy both equations so that:
(Lx + ksina - Cx)2 + (Ly + kcosa - Cy)2 = r2
This equation needs to be solved for k - everything else has a known numeric value. Since it is a quadratic, in k, there will be 0, 1 or 2 solutions for k depending on the line not intersecting the circle at all, being tangent to it and going through the circle.
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