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.

### Answer

AB = 8 cm

⇒ AM = 4 cm

∴ OM = √(5^{2} – 4^{2}) = 3 cm

Let AP = y cm, PM = x cm

∴ ∆OPA is a right angle triangle

∴ OP^{2} = OA^{2} + AP^{2}

(x + 3)^{2} = y^{2} + 25

⇒ x2 + 9 + 6x = y^{2} + 25 ...(i)

Also, x^{2} + 4^{2} = y^{2} ...(ii)

⇒ x^{2} + 6x + 9 = x^{2} + 16 + 25

⇒ 6x = 32

⇒ x = 32/6 = 16/3 cm

∴ y^{2} = x^{2} + 16 = 256/9 + 16 = 400/9

⇒ y = 20/3 cm