Consider ΔPOA
OA = 6 cm …radius of the outer circle
PA = 10 cm …given
∠OAP = 90° …radius is perpendicular to the tangent
Hence ΔPOA is right-angled triangle
Using Pythagoras
⇒ OA2 + AP2 = OP2
⇒ 62 + 102 = OP2
⇒ 36 + 100 = OP2
⇒ OP2 = 136 …(i)
Consider ΔPBO
OB = 4 cm …radius of inner circle
∠OBP = 90° …radius is perpendicular to the tangent
Hence ΔPOB is right-angled triangle
Using Pythagoras
⇒ OB2 + BP2 = OP2
Using (i)
⇒ 42 + BP2 = 136
⇒ 16 + BP2 = 136
⇒ BP2 = 120
⇒ BP = 10.9 cm
Hence length of PB is 10.9 cm