Two concentric circles are of radii 10 cm, and 6cm Find the length of the chord of the larger circle which touches the smaller circle.

Let the two concentric circles have the centre O and let AB be the chord of an outer circle whose length is D and which will also be tangent to the inner circle at point D because it is given that the chord touches the inner circle.

The radius of inner circle OD = 6 cm and the radius of outer circle OB = 10 cm

In ΔOAB

⇒ OA = OB …radius of outer circle

Hence ΔOAB is isosceles triangle

As radius is perpendicular to tangent OC is perpendicular to AB

OC is altitude from apex and in isosceles triangle the altitude is also the median

Hence AD = DB

Hence AB = 2DB

Consider ΔODB

⇒ ∠ODB = 90° …radius perpendicular to tangent

Using Pythagoras theorem

⇒ OD² + BD² = OB²

⇒ 6² + BD² = 10²

⇒ 36 + BD² = 100

⇒ BD² = 100 – 36

⇒ BD² = 64

⇒ BD = ±8

As length cannot be negative

⇒ BD = 8 cm

⇒ AB = 2 × 8 …since AB = 2BD

⇒ AB = 16 cm