In the adjoining figure, PA and PB are tangents from point P to a circle with centre O. If the radius of the circle is 5 cm and PA ⊥ PB, then the length OP is equal to

1. 5 cm

2. 10 cm

3. 7.5 cm

4. 5√2 cm

Join OA as shown in the figure below:

OA ⊥ PA (∵ radius of a circle and tangent through that point are perpendicular to each other.)

∴ ∠OAP = 90°.

Given, PA ⊥ PB

∴ ∠APB = 90°.

∵ the tangents are equally inclined to the line joining the point and the centre of the circle.

∠APO = $\dfrac{1}{2}$ x ∠APB = 45°.

Since, sum of angles in a triangle = 180°.

In △OAP,

⇒ ∠APO + ∠OAP + ∠AOP = 180°
⇒ 45° + 90° + ∠AOP = 180°
⇒ 135° + ∠AOP = 180°
⇒ ∠AOP = 180° - 135°
⇒ ∠AOP = 45°.

Since, ∠AOP = ∠APO hence, △OAP is an isosceles triangle with OA = AP = 5 cm.

In right angled triangle △OAP,

$OP^2 = OA^2 + AP^2 \\[1em] OP^2 = 5^2 + 5^2 \\[1em] OP^2 = 25 + 25 \\[1em] OP^2 = 50 \\[1em] OP = \sqrt{50} \\[1em] OP = 5\sqrt{2} \text{ cm}$

Hence, Option 4 is the correct option.