Question

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

Answer 1

∵ OQ ⊥ PQ   ...[Radius perpendicular to tangent]

5² = OP² + 13²

QP² = 13² – 5²

QP² = 169 – 25

QP² = 144

QP = 12 cm = PR   ...[Tangent from exterior point are equal]

Also, OC ⊥ AB ⇒ ∠ACO = ∠ACP = 90°

In ΔOQP and ΔACP

∠QPO = ∠CPA   ...[Common angles]

∠OQP = ∠ACP   ...[Each 90°]

∴ ΔOQP ∼ ΔACP   ...[AA Similarity]

⇒ `(OQ)/(AC) = (OP)/(CP) = (OP)/(AP)`

⇒ `5/(AC) = 12/(PC) = 13/(AP)`

Now, `5/(AC) = 13/(AP)`

Now, AC = AQ   ...[Tangent from exterior point are equal]

`5/(AQ) = 13/(AP)`

⇒ `(AP)/(AQ) = 13/5`

⇒ AP = 13x, AQ = 5x

Now, PQ = 12

⇒ AP + AQ = 12

⇒ 13x + 5x = 12

⇒ 18x = 12

⇒ `x = 12/18`

⇒ `x = 2/3`

So, PA = 13x

= `13 xx 2/3`

= `26/3` cm

AQ = AC = 5x

= `5 xx 2/3`

= `10/3`

Similarly, BR = BC = `10/3` cm

Hence, AB = AC + BC

= `10/3 + 10/3`

= `20/3` cm