If a man's wit be wandering, let him study the mathematics. — Francis Bacon
2L solution
Let O be the center of the circle and A the point on it.
- Construct line OA
- Construct the perpendicular to the line OA from point A
3E solution
Let O be the center of the circle and A the point on it.
- Construct a circle with an arbitrary center B on the circle and radius AB, intersecting circle O at C
- Construct the circle with center A and radius AC, intersecting circle B at D
- Construct line AD
Explanation
Solution 2L is easy as a tangent is simply a line perpendicular to a radius at a point on the circle.
Solution 3E is an application of Thales' Theorem. Let DA intersect the bigger circle again at E. ∠ECD is a right angle by Thales' theorem. Let EC intersect the smaller circle again at F. Since CDAF is cyclic and ∠FCD = 90°, ∠FAD = 90° and it passes through the center. So, DE is a tangent.