Trigonometry problem for collision detection and reflection

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darkdave3000
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A particle moves from point 0 to point 2. Both positions are known. The center of the circle and it's radius is also known. I am trying to work out the position of 1 where the particle strikes the circle.

This is for a 2D astronomy simulator to work out where a particle will strike a 2D representation of Earth. I've exhausted my brain and I am hoping you guys can lend a hand. I've tried using SecTheta but its an imperfect solution, just like Tan Theta.

Purple lines and points are known magnitudes and positions and any other colors are unknowns.

The way the simulator works is it uses the Euler method to increment the position of a moving object, so the particle will move from say point 0 to point 2 between a computer cycle. The time passed typically is 0.1 seconds.

Please help. Kind Regards, David
And may the force be with u.
 

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darkdave3000 said:
A particle moves from point 0 to point 2. Both positions are known. The center of the circle and it's radius is also known. I am trying to work out the position of 1 where the particle strikes the circle.

This is for a 2D astronomy simulator to work out where a particle will strike a 2D representation of Earth. I've exhausted my brain and I am hoping you guys can lend a hand. I've tried using SecTheta but its an imperfect solution, just like Tan Theta.

Purple lines and points are known magnitudes and positions and any other colors are unknowns.

The way the simulator works is it uses the Euler method to increment the position of a moving object, so the particle will move from say point 0 to point 2 between a computer cycle. The time passed typically is 0.1 seconds.

Please help. Kind Regards, David
And may the force be with u.
How about this?
In the called-out triangle in your drawing, you know the two longer sides and the side opposite the longer one. Use the Law of Sines to get the angle at point 1. Once that angle is known, you can get the third angle (at the bottom end of your drawing), and use the Law of Cosines to get the length of the short side of your triangle.

With all angles and sides of the triangle known, you can write the coordinates of point 1 as (x, y). Point 1 is on a circle of known radius, from which you can derive an equation in x and y. Point 1 is also a known distance from Point 2, which gives you another equation in x and y. Solve these two equations simultaneously to find the coordinates of Point 1.
 
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Mark44 said:
How about this?
In the called-out triangle in your drawing, you know the two longer sides and the side opposite the longer one. Use the Law of Sines to get the angle at point 1. Once that angle is known, you can get the third angle (at the bottom end of your drawing), and use the Law of Cosines to get the length of the short side of your triangle.

With all angles and sides of the triangle known, you can write the coordinates of point 1 as (x, y). Point 1 is on a circle of known radius, from which you can derive an equation in x and y. Point 1 is also a known distance from Point 2, which gives you another equation in x and y. Solve these two equations simultaneously to find the coordinates of Point 1.

Sounds solid! Ok I will do it! There is no try with the force :)