Figure 12. Determination of the equation of a ray that passes through a point B. Use the fact that the origin O is also a point on the ray. Connect these two known points with line OB. Use the fact that the ray is a circle. Hence the center lies on the perpendicular bisector AE of line OB. Use the fact that this center also lies on line JE on which the velocity would be zero. Thus the center is at the intersection E of these two lines. The radius is given by line EB. The raypath is the circle with this center and radius.

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