Question

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact to the centre.

Hint:

In circle geometry, always remember: the tangent is perpendicular to the radius at the point of contact. This property simplifies most tangent-related proofs.

Answer 1

Step 1: Construction and given information.
Let a circle with centre \( O \) have tangents \( PA \) and \( PB \) drawn from an external point \( P \). Points \( A \) and \( B \) are the points of contact of the tangents.
Step 2: Join \( OA \) and \( OB \).
These are radii drawn to the points of contact, and hence each makes a right angle with the tangents: \[ \angle OAP = \angle OBP = 90^\circ \]
Step 3: In quadrilateral \( OAPB \):
Sum of all interior angles of a quadrilateral is \( 360^\circ \). \[ \angle OAP + \angle OBP + \angle AOB + \angle APB = 360^\circ \] \[ 90^\circ + 90^\circ + \angle AOB + \angle APB = 360^\circ \] \[ \angle AOB + \angle APB = 180^\circ \]
Step 4: Conclusion.
\[ \boxed{\text{The angle between the two tangents (}\angle APB\text{) is supplementary to } \angle AOB.} \]