How do I find two points at an angle on opposite sides of a circle?

Question

With reference to my previous question:

I'm using micklh's answer, which works perfectly. However, it calculates the gradient based on two points, a starting point and an ending point. I want to change that such that I can give it a center point, an angle in degrees, and a length.

If you look at the picture from the other question:

You can sort of discern where the center point is, the x,y of starting/ending locations become the dots based on the length of the gradient.

I assume I would have to:

1. Take the center location
2. Apply the angle + length
3. Gives me my ending x,y
4. Take the center location
5. Calculate the reverse angle
6. Apply the reverse angle + length
7. Gives me my starting x,y
8. Plug these starting/ending positions to the answer from the other question.

Am I correct? How would I do those steps?

Answer 1

You might want to briefly refresh your trigonometry (maybe by reading the overview at Wikipedia) if you're stuck.

Your problem looks like this:

Keeping that picture in mind, we can take essentially the same steps you described, but in a more mathematical form:

As inputs, you have a centre point p0 (red) and an angle (blue) and a length (orange).
As outputs, you want two points p1 and p2 (in black), representing the points that are length/2 away from p0 on its directly opposite sides.

The gray circle is just for emphasis that trigonometry applies.

Basic trigonometry says that to get the x value of the orange vector (p0 to p1), we can do r * sin(a) and for its y, do r * cos(a).

Hence to find p1, add the orange vector to p0. For p2, subtract the orange vector from p0.

You can then use p1 and p2 to draw your gradient.