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 \(r = 13\).