Q9 of 46 Page 1

Tangents PA and PB are drawn from an external point P to two concentric circle with centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, then find the length of BP.

Given that: AP = 15 cm, AO = 8 cm and OB = 5cm

The diagram is given below:



Since, AP and BP are tangents to the outer and inner circle respectively, this implies that ∠OAP = ∠OBP = 90°.


To find PB, we can first find OP by right angle triangle AOP.


So in ∆AOP, by Pythagoras theorem


(hypotenuse)2 = (base)2 + (perpendicular)2


⇒ (OP)2 = (AO)2 + (AP)2


⇒ (OP)2 = (8)2 + (15)2


⇒ (OP)2 = 64 + 225 = 289


⇒ OP = √289


⇒ OP = 17


Thus, we get OP = 17 cm.


Now in ∆BOP, by Pythagoras theorem


(hypotenuse)2 = (base)2 + (perpendicular)2


⇒ (OP)2 = (OB)2 + (BP)2


⇒ (BP)2 = (OP)2 – (OB)2


⇒ (BP)2 = (17)2 – (5)2


⇒ (BP)2 = 289 – 25 = 264


⇒ BP = √264


⇒ BP = 2√66


Hence, we get BP = 2√66.


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