Lets say you have 2 points A and B. A is given as (x, y) and B is a point that revolves on a circle. You are given the radius (r), the initial angle (i), and the velocity that it travels at (v_b). You are also given the velocity that A travels at (v_a). The center of the circle is (0,0) and all points relate it that point. You want to find the shortest amount of time it takes for A to reach B. To do it in the shortest amound of time, A must travel in a straight line. In order to do that, you must know where B will be in the same amount of time it takes A to get there. I've created 2 equations for distance that relate to time. d(t) = sqrt( (x - rcos(at + i))^2 + (y - rsin(at + i))^2 ) d(t) = (v_a)t These combine to make (v_a)t = sqrt( (x - rcos(at + i))^2 + (y - rsin(at + i))^2 ) *note: sqrt = square root d = distance (variable) t = time (variable) x = x-coordinate of A (constant) y = y-coordinate of A (constant) r = radius (distance from (0,0) to B) (constant) i = initial angle (constant) v_a = velocity of A (constant) v_b = velocity of B (constant) a = angular velocity (v_b / r) (constant) The thing that is wrong with this equation is that you cannot isolate t, but you can use it to approximate t. I've tried finding d'(t) and (d^2)''(t) to eliminate t from one of the functions (I'm not sure they are functions). d'(t) = ( axrsin(at + i) - ayrcos(at + i) ) / sqrt( (x - rcos(at + i))^2 + (y - rsin(at + i))^2 ) d'(t) = v_a (d^2)''(t) = 2(a^2)xrcos(at + i) + 2(a^2)yrsin(at + i) (d^2)''(t) = 2(v_a)^2 But, these functions don't help in finding t. The main purpose is to find the shorest amount of time it takes for A to get to B. So, there's got to be another method, right?