Question

O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.

Answer 1

In the right ∆OTP,

PT² = OT² – OP²

= 13² – 5²

= 169 – 25

= 144

PT = `sqrt(144)` = 12 cm

Since lengths of tangent drawn from a point to circle are equal.

∴ AP = AE = x

AT = PT – AP

= (12 – x) cm

Since AB is the tangent to the circle E.

∴ OE ⊥ AB

∠OEA = 90°

∠AET = 90°

In ∆AET, AT² = AE² + ET²

In the right ∆AET,

AT² = AE² + ET²

(12 – x)² = x² + (13 – 5)²

144 – 24x + x² = x² + 64

24x = 80 ⇒ x = `80/24 = 20/6 = 10/3`

BE = `10/3 "cm"`

AB = AE + BE

= `10/3 + 10/3 = 20/3`

Length of AB = `20/3` cm