Question

Two circles touch externally. The sum of their areas is $130\mathrm{\pi }\mathrm{sq}-\mathrm{cm}$ and the distance between their centres is $14\mathrm{cm}$. Find the radii of the circles.

Answer 1

Radius of a circle:

Distance between the center and the circumference of the circle is called the radius of the circle.

Step-1 : Formation of equation:

Consider two circles$\mathrm{A}\mathrm{and}\mathrm{B}$ with radius $\mathrm{r}\mathrm{and}\mathrm{R}.$

Area of circle$={\mathrm{\pi r}}^2$ square units

Area of circle $A$ $={\mathrm{\pi r}}^2$

Area of circle $B$ $={\mathrm{\pi R}}^2$

The sum of areas of two circles is given as $130\mathrm{\pi }sq-cm.$

$$\begin{gathered} \Rightarrow {\mathrm{\pi r}}^2+{\mathrm{\pi R}}^2=130\mathrm{\pi } \\ \Rightarrow {\mathrm{r}}^2+{\mathrm{R}}^2=130...\left(1\right) \end{gathered}$$

The distance between the center of two circles is given as $14cm$.

Thus, the distance between center of the circle is the sum of the radii of two circle.

$\Rightarrow \mathrm{r}+\mathrm{R}=14...\left(2\right)$

Step-2 : Calculating the radius of circle A:

Squaring equation $\left(2\right)$ on both sides.

$$\begin{gathered} \Rightarrow {\left(\mathrm{r}+\mathrm{R}\right)}^2={14}^2\mathbf{[}{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ab}\mathbf{]} \\ \Rightarrow {\mathrm{r}}^2+{\mathrm{R}}^2+2\mathrm{rR}=196 \\ \Rightarrow 130+2\mathrm{rR}=196 \\ \Rightarrow 2\mathrm{rR}=196-130 \\ \Rightarrow 2\mathrm{rR}=66 \\ \Rightarrow \mathrm{rR}=33 \end{gathered}$$

By using the formula ${\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\left(a+b\right)}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ab}$

$$\begin{gathered} \Rightarrow {(\mathrm{r}-\mathrm{R})}^2={\left(\mathrm{r}+\mathrm{R}\right)}^2-4\mathrm{rR} \\ \Rightarrow {(\mathrm{r}-\mathrm{R})}^2={14}^2-4\left(33\right) \\ \Rightarrow {(\mathrm{r}-\mathrm{R})}^2=196-132 \\ \Rightarrow {(\mathrm{r}-\mathrm{R})}^2=64 \\ \Rightarrow \mathrm{r}-\mathrm{R}={64}^2 \\ \Rightarrow \mathrm{r}-\mathrm{R}=8...\left(3\right) \end{gathered}$$

Solving equations $\left(2\right)$ and $\left(3\right)$.

$$\begin{gathered} \mathrm{r}+\mathrm{R}=14...\left(2\right) \\ \mathrm{r}-\mathrm{R}=8...\left(3\right) \\ \Rightarrow 2\mathrm{r}=22\mathbf{\left(}\mathbf{Adding}\mathbf{\right)} \\ \Rightarrow \mathrm{r}=11 \end{gathered}$$

Thus, the radius of the circle $A$is $11cm$.

Step-3 : Calculating the radius of circle B:

Substitute the value r in equation $\left(2\right)$.

$$\begin{gathered} 11+\mathrm{R}=14 \\ \Rightarrow \mathrm{R}=14-11 \\ \Rightarrow \mathrm{R}=3 \end{gathered}$$

Thus, the radius of the circle B is $3$ cm.

Hence, the radii of the circles $\mathrm{A}\mathrm{and}\mathrm{B}$ are $11cm$ and $3cm$.