question_answer

PA and PB are two tangents drawn from and external point P to a circle with centre O where the points A and B are the points of contact. The quadrilateral OAPB must be

A)  a rectangle                   

B)  a rhombus

C)  a square                      

D)  concyclic

Correct Answer: D

Solution :

OA \[\bot \] AP and OB \[\bot \] BP \[\angle \]OAP = \[90{}^\circ \] and \[\angle \]OBP = \[90{}^\circ \] \[\therefore \] \[\angle \]OAP + \[\angle \]OBP = \[90{}^\circ \] +\[90{}^\circ \]= \[180{}^\circ \] In quadrilateral OAPB, \[\angle \]OAP + \[\angle \]APB + \[\angle \]AOB+\[\angle \]OBP=\[360{}^\circ \] \[\Rightarrow \]\[\angle \]APB + \[\angle \]AOB = \[180{}^\circ \] \[\therefore \] The quadrilateral will be cyclic.