Consider a unit circle with center $O$. Let $P$ be a point outside the circle such that the two line segments passing through $P$ and tangent to the circle form an angle of $60^\circ$. Compute the length of $OP$.
1. Let \( O \) be the center of the unit circle, and let \( P \) be a point outside the circle such that the two line segments passing through \( P \) and tangent to the circle form an angle of \( 60^\circ \). Let \( A \) and \( B \) be the points of tangency of the two line segments from \( P \) to the circle.

2. Since \( PA \) and \( PB \) are tangents to the circle from point \( P \), we know that \( PA = PB \). Also, the radii \( OA \) and \( OB \) are perpendicular to the tangents at the points of tangency, so \( \angle OAP = \angle OBP = 90^\circ \).

3. The angle between the tangents \( PA \) and \( PB \) is given as \( 60^\circ \). Therefore, the angle \( \angle APB = 60^\circ \).

4. Since \( PA = PB \), triangle \( \triangle APB \) is isosceles with \( \angle PAB = \angle PBA \). Let \( \angle PAB = \angle PBA = \theta \). Then, we have:
   \[
   \theta + \theta + 60^\circ = 180^\circ \implies 2\theta = 120^\circ \implies \theta = 60^\circ
   \]

5. Therefore, \( \triangle OAP \) and \( \triangle OBP \) are right triangles with \( \angle OAP = \angle OBP = 90^\circ \) and \( \angle AOP = \angle BOP = 30^\circ \).

6. Consider the right triangle \( \triangle OAP \). Since \( \angle OAP = 90^\circ \) and \( \angle AOP = 30^\circ \), \( \triangle OAP \) is a \( 30-60-90 \) triangle.

7. In a \( 30-60-90 \) triangle, the ratio of the sides opposite the \( 30^\circ \), \( 60^\circ \), and \( 90^\circ \) angles are \( 1 : \sqrt{3} : 2 \). Since \( OA = 1 \) (the radius of the unit circle), we have:
   \[
   OP = 2 \times OA = 2 \times 1 = 2
   \]

8. Therefore, the length of \( OP \) is \( 2 \).

The final answer is \( \boxed{2} \)