Let O be the common centre of two concentric circles and let AB be a chord of larger circle touching the smaller circle at P. Join OP.
Since OP is the radius of the smaller circle and AB is tangent to this circle at P,
OP ⊥ AB
We know that the perpendicular drawn from the centre of a circle to any chord of
the circle bisects the chord.
Therefore, AP = BP
In right DAPO we have
Hence, the length of the chord of the larger circle which touches the smaller circle is 2 √39 cm.
