Question

Question 10
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-segment joining the points of contact at the centre.

Answer 1

Consider a circle with centre O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends $\angle$AOB at center O of the circle.
It can be observed that,
OA $\perp$ PA
$\therefore \angle OAP={90}^{\circ }$
Similarly, OB $\perp$ PB
$\therefore \angle OBP={90}^{\circ }$
In quadrilateral OAPB,
Sum of all interior angles = ${360}^{\circ }$
$\angle OAP+\angle APB+\angle PBO+\angle BOA={360}^{\circ }$
$\Rightarrow {90}^{\circ }+\angle APB+{90}^{\circ }+\angle BOA={360}^{\circ }$
$\Rightarrow \angle APB+\angle BOA={180}^{\circ }$
$\therefore$ The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.