When two tangents, PA and PB, are drawn to a circle from a point P outside the circle, and the angle between them is 90°,what is the distance from P to the center (O) of the circle, if the radius is 5 cm?

Correct — B. Each radius meets its tangent at 90°, so quadrilateral OAPB has right angles at A and B.

Given: ∠APB = 90°
Angle sum of a quadrilateral: ∠AOB = 360° − 90° − 90° − 90° = 90°
All four angles 90° and adjacent sides OA = OB = 5 cm: OAPB is a square of side 5 cm

OP is its diagonal: OP² = 5² + 5² = 50
Take the root: OP = 5√2 cm → option (b)

Why the others are wrong

• (a)
  5 cm — 5 cm is the radius itself. A point 5 cm from O lies on the circle, and the question puts P outside it.
• (c)
  10 cm — 10 cm is 2 × the radius. Then sin ∠APO = 5⁄10, so ∠APO = 30° and the tangents meet at 60°, not 90°.
• (d)
  5√3 cm — 5√3 cm gives PA = √(75 − 25) = 5√2 cm. But with the tangents at 90°, △OAP is isosceles, so PA must equal the radius, 5 cm.

Two facts about tangents from an external point P settle this item. The radius to each point of contact is perpendicular to its tangent, and OP bisects the angle between the two tangents.

So ∠APO is half of ∠APB. With ∠APB = 90°, ∠APO = 45°, △OAP is right isosceles, and OP = r√2.

Key facts

• Tangents from an external point are equal: PA = PB.
• OP bisects the angle between the two tangents.
• ∠APB + ∠AOB = 180°, because the angles at A and B are both 90°.
• Tangents meeting at 90° make OAPB a square, so OP = r√2.

Common traps

• Taking OP as a side of the square (5 cm) rather than its diagonal.
• Assuming OP = 2r for any pair of tangents. That holds only when they meet at 60°.

No directly related past PYQ was found.