Review question

Ref: R7032

## Solution

The sum of the circumferences of two circles is $\quantity{60\pi}{cm}$. and the sum of their areas is $\quantity{458\pi}{sq.cm.}$ Calculate the radii of the circles.

If a circle has radius $\quantity{x}{cm}$, then its circumference is $\quantity{2\pi x}{cm}$ and its area is $\quantity{\pi x^2}{sq.cm.}$

Let us denote the radii of the two circles in the question by $\quantity{r}{cm}$ and $\quantity{R}{cm}$. The information in the question leads to two equations: \begin{align*} 2\pi r + 2\pi R &= 60 \pi, \\ \pi r^2 + \pi R^2 &= 458 \pi. \end{align*} By dividing the first equation through by $2\pi$ and the second by $\pi$, these equations become \begin{align*} r + R &= 30, \\ r^2 + R^2 &= 458. \end{align*}

If $r = R$, then $r = R = 15$, and the second equation does not hold. So $r \neq R$, and we can say $R > r$.

The first equation implies that $r = 30 - R$, so substituting this into the second, we have that \begin{align*} (30 - R)^2 + R^2 = 458 &\implies 900 - 60R + R^2 + R^2 = 458 \\ &\implies 2R^2 - 60R + 442 = 0 \\ &\implies R^2 - 30R + 221 = 0. \end{align*}

(Note that since the equations for $r$ and $R$ are symmetrical, $r$ must be a solution to $r^2 - 30r + 221 = 0$.)

By applying the quadratic formula, we know that \begin{align*} R &= \frac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times 221}}{2 \times 1} \\ &= \frac{30 \pm \sqrt{900 - 884}}{2} \\ &= \frac{30 \pm \sqrt{16}}{2} \\ &= \frac{30 \pm 4}{2} \\ &= 17 \quad \text{or} \quad 13. \end{align*}

So $R = 17$ and $r = 13$.