Question

State True or False.

PA and PB are tangents to the circle with centre O from an external point P, touching the circle at A and B respectively. Then the quadrilateral AOBP is cyclic.

• A. Point of secant
• B. Point of contact

Answer 1

The correct option is A

Point of secant

Given PA and PB are tangents to the circle with centre O from an external point P.

We know that the tangent at any point of a circle is perpendicular to radius at the point of contact.

$\therefore PA\perp OA,i.e.,\angle OAP={90}^{\circ }$ . . . (i)
and $PB\perp OB,i.e.,\angle OBP={90}^{\circ }$ . .. (ii)

Now, the sum of all the angles of a quadrilateral is ${360}^{\circ }$

$\therefore \angle AOB+\angle OAP+\angle APB+\angle OBP={360}^{\circ }$
$\Rightarrow \angle AOB+\angle APB={180}^{\circ }$ [using (i) and (ii)]

$\therefore$ quadrilateral OAPB is cyclic [since both pairs of opposite angles have the sum ${180}^{\circ }$].