Q) PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If OP = 13 cm, then find the length AB and PA.

(Q 33 – 30/5/2 – CBSE 2026 Question Paper)

Ans:

It is given that OP = 13 cm, OQ = OR 5 cm (radii)

PQ, PR & AB are tangents

Step 1: Radius OQ is Ʇ tangent PQ,

∴ ∠ QOP is 90 ⁰

Now by Pythagoras theorem in Δ OQP:

OP ² = OQ ² + PQ ²

It is given Radius OQ = 5 cm, OP = 13 cm

∴ 13 ² = 5 ² + PQ ²

∴ PQ = 12                  (by Pythagoras triplets)

Step 2: Let’s consider AC = x

Given that AB is tangent to circle,

∴ AC is tangent to circle

Now, AQ & AC are tangents from external point A

∴ AQ = AC = x

∴ PA = PQ – AQ = 12 – x

Also OP = OC + CP

∴ CP = OP – OC

∴ CP = 13 – 5 = 8 cm

Step 3: Since radius OC is Ʇ tangent AC,

∴ ∠ ACO is 90 ⁰

Now, By Pythagoras theorem in right angled Δ ACP

∴ PA ² = AC ² + CP ²

By substituting values as calculated above, we get:

∴ (12 – x) ² = x ² + 8 ²

∴ 144 + x ² – 24 x = x ² + 8 ²

∴ 144 – 24 x = 64

∴ – 24 x = 64 – 144 = – 80

∴ x =

∴ AC = x = = 3.33 cm

(i) Length of AB:

∵ AB = 2 x AC (OC is Ʇ bisector of AB)

∴ AB = 2 x = 6.67 cm

(ii) Length of PA:

∵ PA = 12 – x

∴ PA = 12 –

∴ PA = = 8.67 cm

Therefore, the lengths of AB and PA are 6.67 cm and 8.67 cm.

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