Question

In the following figure, PA and PB are tangents from a point P to a circle with centre O. Then the quadrilateral OAPB must be a

• A. Square
• B. Rhombus
• C. Cyclic quadrilateral
• D. Parallelogram

Answer 1

The correct option is C Cyclic quadrilateral

Since tangent and radius intersect at right angle,

So,

∠OAP = ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90° = 180°

Which are opposite angles of quadrilateral OAPB

So the sum of remaining two angles is also 180°

Therefore Quadrilateral OAPB is cyclic Quadrilateral.
So, option c is correct.