In figure, AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP.
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AB = 8 cm
⇒ AM = 4 cm
∴ OM = √(52 – 42) = 3 cm
Let AP = y cm, PM = x cm
∴ ∆OPA is a right angle triangle
∴ OP2 = OA2 + AP2
(x + 3)2 = y2 + 25
⇒ x2 + 9 + 6x = y2 + 25 ...(i)
Also, x2 + 42 = y2 ...(ii)
⇒ x2 + 6x + 9 = x2 + 16 + 25
⇒ 6x = 32
⇒ x = 32/6 = 16/3 cm
∴ y2 = x2 + 16 = 256/9 + 16 = 400/9
⇒ y = 20/3 cm
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