The radii of two circles are 19 cm and 9 cm respectively.

Solution:

Let the radius of circles be $r \mathrm{~cm}, r_{1} \mathrm{~cm}$ and $r_{2} \mathrm{~cm}$ respectively. Then their circumferences are $C=2 \pi r \mathrm{~cm}, C_{1}=2 \pi r_{1} \mathrm{~cm}$ and $C_{2}=2 \pi r_{2} \mathrm{~cm}$ respectively.

It is given that,

Circumference $C$ of circle $=$ Circumference $C_{1}$ of circle $+$ Circumference $C_{2}$ of circle

$2 \pi r=2 \pi r_{1}+2 \pi r_{2}$

$2 \pi r=2 \pi\left(r_{1}+r_{2}\right)$

$r=r_{1}^{*}+r_{2}$

We have, $r_{1}=19 \mathrm{~cm}$ and $r_{2}=9 \mathrm{~cm}$

Substituting the values of $r_{1}, r_{2}$

$r=19+9$

$r=28 \mathrm{~cm}$

Hence the radius of the circle is $28 \mathrm{~cm}$.

We know that the area $A$ of circle is

$A=\pi r^{2}$

Substituting the value of r

$A=\frac{22}{7} \times 28 \times 28$

$=2464 \mathrm{~cm}^{2}$

Hence the area of the circle is $2464 \mathrm{~cm}^{2}$.