Two concentric circles are of radii 5 cm and 3 cm.

In fig. the two concentric circles have their centre at O. The radius of the larger circle is 5 cm and that of the smaller circle is 3 cm.

AB is a chord of the larger circle and it touches the smaller circle at P.

Join $\mathrm{OA}, \mathrm{OB}$ and $\mathrm{OP}$.

$\mathrm{OP}=3 \mathrm{~cm}$

and $\mathrm{OP} \perp \mathrm{AB}$,

i.e., $\angle \mathrm{OPA}=$

$\angle \mathrm{OPB}=90^{\circ}$

$\Rightarrow \quad \Delta \mathrm{OAP} \cong \Delta \mathrm{OBP} \quad(\mathrm{RHS}$ congruence $)$

$\Rightarrow \quad \mathrm{AP}=\mathrm{BP}=\frac{\mathbf{1}}{\mathbf{2}} \mathrm{AB}$ or $\mathrm{AB}=2 \mathrm{AP}$

By Pythagoras theorem,

$\mathrm{OA}^{2}=\mathrm{AP}^{2}+\mathrm{OP}^{2}$

$\Rightarrow \quad(5)^{2}=A P^{2}+(3)^{2}$

$\mathrm{AP}^{2}=25-9=16$

$\Rightarrow \quad \mathrm{AP}=4 \mathrm{~cm}$

$\Rightarrow \quad \mathrm{AB}=2 \times 4 \mathrm{~cm}=8 \mathrm{~cm}$

Now, $\mathrm{OA}=\mathrm{OB}=5 \mathrm{~cm}$