Q10 of 31 Page 233

P lies in the exterior of ⨀(O, 5) such that OP = 13. Two tangents are drawn to the circle which touch the circle in A and B. Find AB.

Given P lies in the exterior of circle (O, 5) such that OP = 13.


Let OR = x.



Consider ΔOAP where ∠A = 90°,


By Pythagoras Theorem,


⇒ OP2 = OA2 + AP2


⇒ 132 = 52 + AP2


⇒ AP2 = 169 – 25 = 144


∴ AP = 12


Consider ΔORA where ∠R = 90°,


⇒ AR2 = OA2 – OR2


⇒ AR2 = 52 – x2 = 25 – x2 … (1)


Consider ΔARP where ∠R = 90°,


⇒ AR2 = AP2 – PR2


= (12)2 – (13 – x)2


= 144 – (13 + x2 – 26x)


= – x2 + 26x + 131 … (2)


From (1) and (2),


⇒ 25 – x2 = – x2 + 26x + 131


⇒ 26x = 50


⇒ x =


From (1),


⇒ AR2 = 25 – x2


= 25 – () 2


= 25 –


=


=


∴ AR =


But AB = 2AR,


⇒ AB = 2 ()


∴ AB = = 9.23


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