Question

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Hint:

Remember: Perpendicularity is equivalent to the shortest distance property. This theorem is the foundation for solving most circle-tangent problems.

Answer 1

Step 1: Understanding the Concept:
We need to prove that if \( XY \) is a tangent to a circle with center \( O \) at point \( P \), then \( OP \perp XY \).
Step 2: Key Formula or Approach:
We will use the property that a tangent at any point of a circle is perpendicular to the radius drawn to the point of contact.
Step 3: Detailed Explanation:
1. Let \( XY \) be a tangent to the circle at point \( P \), and \( O \) be the center of the circle.
2. By the property of tangents, the line drawn from the center of the circle to the point of tangency is perpendicular to the tangent. Thus, \( OP \perp XY \).
3. Suppose for the sake of contradiction that \( OP \) is not perpendicular to \( XY \). Then, we would have an angle \( \theta \) between \( OP \) and \( XY \) that is not \( 90^\circ \).
4. If \( \theta \neq 90^\circ \), then by geometry, we would be able to draw a line from \( O \) to the point where the perpendicular from \( O \) meets \( XY \), which would form a right triangle with a shorter distance than \( OP \).
5. This contradicts the assumption that \( OP \) is the shortest distance from the center \( O \) to the line \( XY \).
6. Therefore, \( OP \perp XY \).
Step 4: Final Answer:
The radius at the point of tangency is always perpendicular to the tangent.