Rotate a circle around another circle

Question

Short question: Given a point P and a line segment L, how do I find the point (or points) on L that are exactly X distance from P, if it guaranteed that there is such a point?

The longer way to ask this question is with an image. Given two circles, one static and one dynamic, if you move the dynamic one towards the static one in a straight line, it's pretty easy to determine the point of contact (see 1, the green dot).

Now, if you move the dynamic circle towards the static circle at an angle, determining the point of contact is much more difficult, (see 2, the purple dot). That part I already have done. What I want to do is, after determining the point of contact, decrease the angle and determine the new point of contact (see 3, 4, the red dot).

In #4, you can see the angle is decreased by less than half, and the new point of contact is half-way between the straight-line point and the original point. In #7, you can see the angle is bisected, but the new point of contact moves much farther than half way back towards the straight-line point.

In my case, I always want to decrease the angle to 5/6ths its original value, but the original angle and distance between the circles are variable. The circles are all the same radius. The actual data I need after decreasing the angle is the vector between the new center of the dynamic circle and the static circle, that is, the blue line in 3, 4, 6, and 7, if that makes the calculation any easier.

So far, I know I have to move the dynamic circle along the line that the purple circle is a center of, towards the center of the static circle. Then the circle has to move directly back towards the original position of the dynamic circle. The hard part is knowing exactly how far back it has to move so that it's just touching the other circle.

Answer 1

Draw a circle with the same centre as the stationary circle and the radius of the sum of both radii. There are two intersections with the translation line of the moving circle's centre. The place of the moving circle's center at the time of contact is the closer of those two intersections.

Answer 2

To answer your short question, if you are on the Cartesian plane, then find the equation of the line L is sitting on (given the two endpoints of L, this is simple). Find the equation of the perpendicular to said line, which passes through P (this is done by taking the negative inverse of the slope, plugging in P's x and y values, and solving for the intercept). Then find the point where the two perpendicular lines intersect by using their equations as a single system of equations (with x's and y's equal). Then find the distance between the point of intersection and the point P, which is one leg of a triangle. Finally, with that distance and the distance X you are given, use Pythagorean theorem to find the distance of the other leg of the triangle. Now the point you are looking for is a point on L, and also on the line on which L sits. So using the distance you just obtained, the intersection point you had found before, and the equation of L's line, you can find the desired point's coordinates. There can only be a maximum of 2 such points, so all you have to test for is whether the coordinates of the points found are actually on L, or beyond L but still on its line. Sorry for the long answer and if you wanted a geometric explanation rather than an algebraic one.

Answer 3

Let the ends of your segment be A and B, and the center of your stationary circle be C. Let the radius of both circles be r. Let the center of the moving circle at the moment of collision be D. We have a triangle ACD, of which we know: the distance AC, because it is constant, the angle DAC, because that's what you are changing, and the distance CD, which is exactly 2r. Theoretically, two sides and angle should let you get all the rest of a triangle...