text: The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm.
  
  Question:
  
  The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumference of the circles.
  
  Solution:
  
  Let the radii of the two circles be r1 cm and r2 cm.
  Now,
  Sum of the radii of the two circles = 7 cm
  
  $r_{1+} r_{2}=7 \quad \ldots(i)$
  
  Difference of the circumferences of the two circles = 88 cm
  
  $\Rightarrow 2 \pi \mathrm{r}_{1}-2 \pi \mathrm{r}_{2}=8$
  
  $\Rightarrow 2 \pi\left(\mathrm{r}_{1}-\mathrm{r}_{2}\right)=8$
  
  $\Rightarrow\left(\mathrm{r}_{1}-\mathrm{r}_{2}\right)=\frac{8}{2 \pi}$
  
  $\Rightarrow \mathrm{r}_{1}-\mathrm{r}_{2}=\frac{8}{2 \times \frac{22}{7}}$
  
  $\Rightarrow \mathrm{r}_{1-} \mathrm{r}_{2}=\frac{8 \times 7}{44}$
  
  $r_{1}-r_{2}=\frac{56}{44}$
  
  $r_{1}-r_{2}=\frac{14}{11} \quad \cdots(i i)$
  
  Adding (i) and (ii), we get:
  
  $2 r_{1}=\frac{91}{11}$
  
  $r_{1}=\frac{91}{22}$
  
  $\therefore$ Circumference of the first circle $=2 \pi r_{1}$
  
  $=2 \times \frac{22}{7} \times \frac{91}{22}$
  
  $=26 \mathrm{~cm}$
  
  Also,
  
  $r_{1}-r_{2}=\frac{14}{11}$
  
  $\frac{91}{22}-r_{2}=\frac{14}{11}$
  
  $\frac{91}{22}-\frac{14}{11}=r_{2}$
  
  $r_{2}=\frac{63}{22}$
  
  $\therefore$ Circumference of the second circle $=2 \pi \mathrm{r}_{2}$
  
  $=2 \times \frac{22}{7} \times \frac{63}{22}$
  
  $=18 \mathrm{~cm}$
  
  Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.