Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in figure. If AP = 15 cm, then find the length of BP.

Given that: OA = 8 cm, OB = 5 cm and AP = 15 cm
To find: BP
Construction: Join OP.
Therefore, tangent to a circle is perpendicular to the radius through the point of contact.
So, OA⊥AP and OB⊥BP
On applying Pythagoras theorem in ΔOAP, we obtain:
(OP)² = (OA)² + (AP)²
⇒ (OP)² = (8)² + (15)²
⇒ (OP)² = 64 + 225
⇒ OP = √289
⇒ OP = 17
Thus, the length of OP is 17 cm.
On applying Pythagoras theorem in ΔOBP, we obtain:
(OP)² = (OB)² + (BP)²
⇒ (17)² = (5)² + (BP)²
⇒ 289 = 25 + (BP)²
⇒ BP = √264
⇒ BP ≈ 16.25
Hence the length of BP is 16.25 cm.