Question

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

Hint:

For tangents from an external point, use the perpendicular property of radius and tangent: \(OA \perp PA\).

Answer 1

Step 1: Draw the figure.
Let \(O\) be the centre of the circle, \(P\) an external point, and \(PA\) and \(PB\) the tangents to the circle touching it at points \(A\) and \(B\).
Step 2: Join \(OA\), \(OB\), and \(OP\).
Since \(OA\) and \(OB\) are radii drawn to the points of contact, they are perpendicular to the tangents: \[ OA \perp PA \quad \text{and} \quad OB \perp PB \]
Step 3: Consider quadrilateral \(OAPB\).
This quadrilateral has two right angles at \(A\) and \(B\).
Step 4: Use the property of cyclic quadrilateral.
The sum of all 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 5: Conclusion.
Hence, the angle between the tangents (\(\angle APB\)) is supplementary to the angle subtended by the line joining the points of contact at the centre (\(\angle AOB\)). \[ \boxed{\angle APB + \angle AOB = 180^\circ} \]