Question

PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at point T. Find the length of TP.

Answer 1

Given radius, OP = OQ = 5 cm
Length of chord, PQ = 4 cm
OT ⊥ PQ,
∴ PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right ΔOPM,
OP² = PM² + OM²
⇒ 52 = 42 + OM²
⇒ OM² = 25 – 16 = 9
Hence OM = 3cm
In right ΔPTM,
PT² = TM² + PM² → (1)
 ∠OPT = 90º [Radius is perpendicular to tangent at point of contact]
In right ΔOPT,
OT2² = PT² + OP² → (2)
From equations (1) and (2), we get
OT² = (TM² + PM²) + OP2²
⇒ (TM + OM)² = (TM² + PM²) + OP²
⇒ TM² + OM² + 2 × TM × OM = TM² + PM² + OP²
⇒ OM² + 2 × TM × OM = PM2 + OP²
⇒ 32 + 2 × TM × 3 = 42 + 52
⇒ 9 + 6TM = 16 + 25
⇒ 6TM = 32
⇒ TM = 32/6 = 16/3
Equation (1) becomes,
 PT² = TM² + PM²
         = (16/3)2 + 42
         = (256/9) + 16 = (256 + 144)/9
         = (400/9) = (20/3)2
⇒ PT = 20/3
∴ the length of tangent PT is (20/3) cm

Answer 2

Step-by-step explanation:

Joint OT. 

Let it meet PQ at the point R. 

Then ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.

[∵TP=TQ= Tangents from T upon the circle]

∴OT⊥PQ

∴OT bisects PQ.

PR=RQ=4 cm

Now, 

OR=OP2−PR2=52−42=3 cm

Now, 

∠TPR+∠RPO=90∘(∵TPO=90∘)

=∠TPR+∠PTR(∵TRP=90∘)

∴∠RPO=∠PTR

∴ Right triangle TRP is similar to the right triangle 

PRO. [By A-A Rule of similar triangles]

∴POTP=RORP⇒5TP=34

⇒TP=320 cm.