Question

Ab is a chord of lenght of 8cm ofva circle of radius 5cm.The tangents to the circle at A and B intersect at P. Find the lenght of AP

Answer 1

Given radius, OP = OQ = 5 cm

Length of chord, PQ = 4 cm
OP ⊥ AB,
∴ AM = MB = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right ΔOAM,
OA2 = AM2 + OM2
⇒ 52 = 42 + OM2
⇒ OM2 = 25 – 16 = 9
Hence OM = 3cm
In right ΔAPM,
AP2 = PM2 + AM2 → (1)
 ∠OAP = 90º [Radius is perpendicular to tangent at point of contact]
In right ΔOAP,
OP2 = AP2 + OA2 → (2)
From equations (1) and (2), we get
OP2 = (PM2 + AM2) + OA2
⇒ (PM + OM)2 = (PM2 + AM2) + OA2
⇒ PM2 + OM2 + 2 × PM × OM = PM2 + AM2 + OA2
⇒ OM2 + 2 × PM × OM = AM2 + OA2
⇒ 32 + 2 × PM × 3 = 42 + 52
⇒ 9 + 6PM = 16 + 25
⇒ 6PM = 32
⇒ PM = 32/6 = 16/3
Equation (1) becomes,
 AP2 = AM2 + PM2 
         = (16/3)2 + 42
         = (256/9) + 16 = (256 + 144)/9
         = (400/9) = (20/3)2
Hence AP = 20/3
Thus, the length of tangent AP is (20/3) cm.