Question

Two circles have circumferences that add up to $12\mathrm{\pi }\mathrm{cm}$and areas that add up to $20\mathrm{\pi }{\mathrm{cm}}^2$. Find the radius of each circle.

Short answer

The radius of the two circles are $2and4$.

Step-by-step solution

Step 1. Given information.

Assume x to be the radius of the first circle and y to be the radius of the second circle.

From the question,

$$\begin{gathered} 2\mathrm{\pi x}+2\mathrm{\pi y}=12\mathrm{\pi }......\left(\mathrm{i}\right) \\ {\mathrm{\pi x}}^2+{\mathrm{\pi y}}^2=20\mathrm{\pi }......\left(\mathrm{ii}\right) \end{gathered}$$

Take out the common factors from equation (i),

$$\begin{gathered} 2\mathrm{\pi }\left(\mathrm{x}+\mathrm{y}\right)=2\mathrm{\pi }\left(6\right) \\ \mathrm{x}+\mathrm{y}=6.......\left(\mathrm{iii}\right) \end{gathered}$$

Take out the common factors from equation (ii),

$$\begin{gathered} \mathrm{\pi }\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)=20\mathrm{\pi } \\ {\mathrm{x}}^2+{\mathrm{y}}^2=20......\left(\mathrm{iv}\right) \end{gathered}$$

Step 2. Solve equation (iii) for x.

Solving equation (iii) for x,

$x=6-y$

Substitute the value of x in equation (iv),

localid="1646849857123" role="math" $$\begin{gathered} {\left(6-y\right)}^2+y^2=20 \\ \left(y-4\right)\left(y-2\right)=0 \\ y=4,2 \end{gathered}$$

Substitute $y=4$in equation (iii),

$$\begin{gathered} x+4=6 \\ x=2 \end{gathered}$$

Substitute $y=2$in equation (iii),

$$\begin{gathered} x+2=6 \\ x=4 \end{gathered}$$