Question

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.

Answer 1

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

see figure , 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 – (169 + x²  – 26x)
= – x² + 26x - 25 … (2)
From (1) and (2),
⇒ 25 – x² = – x² + 26x - 25
⇒ 26x = 50
then, x = 50/26 , put it equation (1).
AR² = 25 - (50/26)² = 25 - (25/13)²
= 25( 1 - 25/169)
= 25 × (169 - 144)/169
= 25 × 144/169 = 3600/169
so, AR = 60/13

∴ AB = 2AR
so, AB = 2 × 60/13 = 120/13 = 9.23

hence, AB = 9.23

Answer 2

Answer:

AB=120/13 units

Step-by-step explanation:

please refer to the attachment!

have a nice day.