Given: Two circles of radii 3 cm and 5 cm with common center.

Let AB be a tangent to the inner/small circle and chord to the larger circle

Let 'P' be the point of contact

Construction: join OP and OB

In △OPB;

∠OPB = 90°

[radius is perpendicular to the tangent]

OP = 3cm and OB = 5cm

Now , OB² = OP²+PB²

[hypotenuse² = Adj.side²+Opp.side², Pythagoras theorem]

² = ²+PB²

PB² = - =

PB = √ = cm

Now , AB = 2×PB

[The perpendicular drawm from the center of the circle to a chord, bisects it]

AB = 2× = cm

The length of the chord of the larger circle which touches the smaller circle is cm