The shortest path between two points in the unit disk that reflects off the circumference is composed of two straight line segments. The problem can be simplified by choosing the coordinate system carefully. We define the

axis along the position of the starting point:

, and define the point of intersection by the angle

from the

axis

, and the final point by a radius

and angle

,

. Then define a symmetry point about

of line

named

.

Then the length of the two line segments is ,

which is minimized by choosing an appropriate

value. This equation can be simplified to

.

The length of the two line segments as a function of

is drawn in the right plot.

There are several simple solutions. If

is 1 or

is 0 or

is 0, the optimal angle

is 0. If

is 1 or

is 0, the optimal angle is

.

Label the origin

. The optimal solution shows that the angle

(from the origin to

to

) is the same as the angle

(from the origin to

to

).

We name these angles

. This can be proved by drawing an ellipse whose foci are

and

. When the ellipse is tangent to the circle, the point of tangency is exactly

.

Since the distance from the origin to

is always 1, we can set up three equalities using the law of sines:

From triangle

:

.

From triangle

:

.

If we mirror the point

about the

axis and label this point

, from triangle

:

.

Simplifying this system of equations results in:

.

Solving this last equation results in a quartic solution that has a closed-form solution with four roots, each of which can be either a clockwise or a counterclockwise rotation

, depending on the sign of

, with

. We evaluate each and select the solution that results in the shortest length path.

Note that the optimal path satisfies the law of reflection off the unit circle, with angle of incidence equal to angle of reflection.