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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.

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