If two concentric circles are of radii 5 cm and 3 cm, then the length of the chord of the larger circle which touches the smaller circle is

Given : $$C_1$$ and $$C_2$$ be the two concentric circles having radius $$r_1=3$$ cm and $$r_2=5$$ cm respectively.

To find : AB = ?

Solution : AB is the the tangent to the circle $$C_1$$, hence $$\angle$$ OPB = $$90^\circ$$

Also, the perpendicular from the centre of a circle to a chord bisects the chord.

Thus, in $$\triangle$$ OPB,

=> $$(PB)^2=(OB)^2-(OP)^2$$

=> $$(PB)^2=(5)^2-(3)^2$$

=> $$(PB)^2=25-9=16$$

=> $$PB=\sqrt{16}=4$$ cm

$$\therefore$$ AB = $$2\times4=8$$ cm

=> Ans - (D)