Point O is the centre of the circle suppose PQ is a chord of length 8 cm of a circle of radius 5 cm the tangent at point P and Q intersect at a point P find TP
Given radius, OP = OQ = 5 cm
Length of chord, PQ = 8 cm
OT ⊥ PQ,
∴ PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right ΔOPM,
OP^2 = PM^2 + OM^2
⇒ 5^2 = 4^2 + OM^2
⇒ OM^2 = 25 – 16 = 9
Hence OM = 3cm
In right ΔPTM,
PT^2 = TM^2 + PM^2 → (1)
∠OPT = 90º [Radius is perpendicular to tangent at point of contact]
In right ΔOPT,
OT^2 = PT^2 + OP^2 → (2)
From equations (1) and (2), we get
OT^2 = (TM^2 + PM^2) + OP^2
⇒ (TM + OM)^2 = (TM^2 + PM^2) + OP^2
⇒ TM^2 + OM^2 + 2 × TM × OM
= TM^2 + PM^2 + OP^2
⇒ OM^2 + 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,
PT^2 = TM^2 + PM^2
= (16/3)^2 + 4^2
= (256/9) + 16 = (256 + 144)/9
= (400/9) = (20/3)2
Hence PT = 20/3
Thus, the length of tangent PT is (20/3) cm.
Given radius, OP = OQ = 5 cm
Length of chord, PQ = 8 cm
OT ⊥ PQ,
∴ PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right ΔOPM,
OP^2 = PM^2 + OM^2
⇒ 5^2 = 4^2 + OM^2
⇒ OM^2 = 25 – 16 = 9
Hence OM = 3cm
In right ΔPTM,
PT^2 = TM^2 + PM^2 → (1)
∠OPT = 90º [Radius is perpendicular to tangent at point of contact]
In right ΔOPT,
OT^2 = PT^2 + OP^2 → (2)
From equations (1) and (2), we get
OT^2 = (TM^2 + PM^2) + OP^2
⇒ (TM + OM)^2 = (TM^2 + PM^2) + OP^2
⇒ TM^2 + OM^2 + 2 × TM × OM
= TM^2 + PM^2 + OP^2
⇒ OM^2 + 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,
PT^2 = TM^2 + PM^2
= (16/3)^2 + 4^2
= (256/9) + 16 = (256 + 144)/9
= (400/9) = (20/3)2
Hence PT = 20/3
Thus, the length of tangent PT is (20/3) cm.
