Q) In the given figure, PA is the tangent to the circle with centre O such that OA = 10cm, AB = 8 cm and AB Ʇ OP . Find the length of PB.

(Q 29 A – 30/3/3 – CBSE 2026 Question Paper)

Ans:

Step 1: Let’s start from Δ OAB,

here, ∠ B = 90 ⁰

∴ Δ OAB is right angled triangle

∴ By Pythagoras theorem in Δ OAB: OA ² = OB ² + AB ²

Given that OA = 10 cm and AB = 8 cm 

∴ (10) ² = OB ² + (8) ²

∴ OB ² = 100 – 64 = 36

∴ OB = 6 cm

Step 2: Next, in Δ ABP,

Here ∠ B = 90 ⁰

∴ Δ ABP is right angled triangle

Again by Pythagoras theorem in Δ ABP: PA ² = AB ² + PB ²

Given that AB = 8 cm

∴ PA ² = (8) ² + PB ²

∴ PA ² = PB ² + 64  ………. (i)

Step 3: Next, in Δ OAP,

Here it is given that PA is tangent to the circle

and OA is the radius of this cicle

Since by cicle’s tangent identity, tangent is Ʇ to the radius

∠ OAP = 90 ⁰

∴ Δ OAP is right angled triangle

Again by Pythagoras theorem in Δ OAP: OP ² = OA ² + PA ²

Given that OA = 10 cm

∴ OP ² = (10) ² + PA ²

∴ OP ² = PA ² + 100 ………. (ii)

Step 4: Substituting the value of PA from equation (i), we get:

∴ OP ² = (PB ² + 64) + 100

∴ OP ² = PB ² + 164 

Step 5: Next, from the diagram, we know that OP = OB + PB

∴ In OP ² = PB ² + 164, we substitute the value of OP

∴ (OB + PB) ² = PB ² + 164 ………. (iii)

Since by algebraic identity, (a + b) ² = a ² + b ² + 2 a b

∴ (OB + PB) ² = OB ² + PB ² + 2 OB . PB

Substituting this value in eqaution (iii), we get:

∴ (OB ² + PB ² + 2 OB . PB) = PB ² + 164

∴ OB ² + 2 OB . PB = 164

Step 6: Substituting value of OB = 6 cm (from step 1)

∴ OB + 2 OB . PB = 164

∴ (6) ² + 2 (6) . PB = 164

∴ 36 + 12 PB = 164

∴ 12 PB = 164 – 36 = 128

∴ PB =

Therefore, the value of PB is .

Please press “Heart” if you liked the solution.