Question

Prove that the lengths of tangents drawn from an external point to a circle are equal and they subtend equal angles at the center.

Hint:

The lengths of tangents drawn from an external point to a circle are always equal, and the angles they subtend at the center are also equal.

Answer 1

Step 1: Define the setup.
Let the center of the circle be \( O \), and let \( P \) be an external point from which two tangents \( PA \) and \( PB \) are drawn to the circle. The points of contact are \( A \) and \( B \). We are asked to prove that \( PA = PB \) and that the angles \( \angle OAP \) and \( \angle OBP \) are equal.
Step 2: Prove the equality of the tangents.
By the property of tangents to a circle, the tangents drawn from an external point to the circle are always equal in length. That is: \[ PA = PB \]
Step 3: Prove the equality of the angles.
In triangles \( OAP \) and \( OBP \), we have:
- \( OA = OB \) (radii of the same circle)
- \( PA = PB \) (tangents from the same external point)
- \( \angle OAP = \angle OBP \) (by the property of tangents)
Thus, by the congruence of the two triangles (\( \triangle OAP \cong \triangle OBP \)), we conclude that: \[ \angle OAP = \angle OBP \]
Step 4: Conclusion.
Therefore, the lengths of the tangents are equal, and the angles subtended by the tangents at the center are also equal.