PQ is a chord of

Given: PQ is a chord and PQ = 8cm

Radius of circle, r = 5cm

To find: Length of TP

Construction: Join OT

Let OT intersect PQ at R

We know that,

Lengths of tangents from external point are equal

So, TP = TQ

In Δ TPQ, TP = TQ

This means TPQ is an isosceles triangle as two sides are equal.

Here, OT is bisector of ∠PTQ,

So, OT ⊥ PQ

[Angle bisector and altitude of an isosceles triangle are same]

So, PR = RQ

[Perpendicular from center to a chord bisects the chord]

In ΔORP, Using Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

(OP)² = (PR)² + (OR)²

⇒ (5)² = (4)² + OR²

⇒ 25 = 16 + OR²

⇒ OR² = 25 – 16

⇒ OR² = 9

⇒ OR = √9

⇒ OR = 3

∴ OR = 3cm …(i)

In ΔPRT, Using Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

(TP)² = (PR)² + (RT)²

⇒ (TP)² = (4)² + RT²

⇒ TP² = 16 + RT² …(ii)

In ΔOPT, Using Pythagoras theorem

(Hypotenuse)² = (Perpendicular)² + (Base)²

(OT)² = (TP)² + (OP)²

⇒ (OT)² = 16 + RT² + (5)² [from (ii)]

⇒ (OR + RT)² = 16 + RT² + 25

⇒ (3 + RT)² = 41 + RT²

⇒ 9 + RT² + 2 × 3 × RT = 41 + RT²

⇒ 9 + 6RT = 41

⇒ 6RT = 41 – 9

⇒ 6RT = 32

Now, substituting the value of RT in eq. (ii), we get

TP² = 16 + RT²