calculating parallel lines in a circle - geometry

Question

I am calculating lines (2 sets of coordinates ) ( the purple and green-blue lines ) that are n perpendicular distance from an original line. (original line is pink ) ( distance is the green arrow )
How do I get the coordinates of the four new points?
I have the coordinates of the 2 original points and their angles. ( pink line )
I need it to work if the lines are vertical, or any other orientation.
Right now I am trying to calculate it by:
1. get new point n distance perpendicular to the two old points
2. find where the circle intersects the new line I have defined.
I feel like there is an easier way.

Answer 1

Similarly to #MBo's answer, let's assume that the center is (0,0) and that your initial two points are:
P0 = (x0, y0) and P1 = (x1, y1)
A point on the line P0P1 has the form:
(x, y) = c(x1 - x0, y1 - y0) + (x0, y0)
for some constant c.
Let (u, v) be the normal to the line P0P1:
(u, v) = (y1 - y0, x1 - x0) / sqrt((x1 - x0)^2 + (y1 - y0)^2)
A point on any of the lines parallel to P0P1 has the form:
(x, y) = c(x1 - x0, y1 - y0) + (x0, y0) +/- (u, v)* n {eq 1}
where n is the perpendicular distance between lines and c is a constant.
What remains here is to find the values of c such that (x,y) is on the circle. But these can be calculated by solving the following two quadratic equations:
(c(x1 - x0) + x0 +/- u*n)^2 + (c(y1 - y0) + y0 +/- v*n)^2 = r^2
where r is the radius. Note that these equations can be written as:
c^2(x1 - x0)^2 + 2c(x1 - x0)*(x0 +/- u*n) + (x0 +/- u*n)^2
+ c^2(y1 - y0)^2 + 2c(y1 - y0)*(y0 +/- v*n) + (y0 +/- v*n)^2 = r^2
or
A*c^2 + B*c + D = 0
where
A = (x1 - x0)^2 + (y1 - y0)^2
B = 2(x1 - x0)*(x0 +/- u*n) + 2(y1 - y0)*(y0 +/- v*n)
D = (x0 +/- u*n)^2 + (y0 +/- v*n)^2 - r^2
which are actually two quadratic equations one for each selection of the +/- signs. The 4 solutions of these two equations will give you the four values of c from which you will get the four points using {eq 1}
UPDATE
Here are the two quadratic equations (I've reused the letters A, B and C but they are different in each case):
A*c^2 + B*c + D = 0 {eq 2}
where
A = (x1 - x0)^2 + (y1 - y0)^2
B = 2(x1 - x0)*(x0 + u*n) + 2(y1 - y0)*(y0 + v*n)
D = (x0 + u*n)^2 + (y0 + v*n)^2 - r^2
A*c^2 + B*c + D = 0 {eq 3}
where
A = (x1 - x0)^2 + (y1 - y0)^2
B = 2(x1 - x0)*(x0 - u*n) + 2(y1 - y0)*(y0 - v*n)
D = (x0 - u*n)^2 + (y0 - v*n)^2 - r^2

Answer 2

Let's circle radius is R, circle center is (0,0) (if not, shift all coordinates to simplify math), first chord end is P0=(x0, y0), second chord end is P1=(x1,y1), unknown new chord end is P=(x,y).
Chord length L is
L = Sqrt((x1-x0)^2 + (y1-y0)^2)
Chord ends lie on the circle, so
x^2 + y^2 = R^2 {1}
Doubled area of triangle PP0P1 might be expressed as product of the base and height and through absolute value of cross product of two edge vectors, so
+/- L * n = (x-x0)*(y-y1)-(x-x1)*(y-y0) = {2}
x*y - x*y1 - x0*y + x0*y1 - x*y + x*y0 + x1*y - x1*y0 =
x * (y0-y1) + y * (x1-x0) + (x0*y1-x1*y0)
Solve system of equation {1} and {2}, find coordinates of new chord ends.
(Up to 4 points - two for +L*n case, two for -L*n case)
I cannot claim though that this method is simpler - {2} is essentially an equation of parallel line, and substitution in {1} is intersection with circle.