The circles touch internally. The sum of their areas is 116π cm² and distance between their centres is 6 cm. Find the radius of larger circle.
10 cm
Since the circles touch internally, the distance between their centres equals the difference of their radii: \(R-r=6\), where \(R\) is the radius of the larger circle.
The sum of the areas is \(\pi R^2+\pi r^2=116\pi\), so \(R^2+r^2=116\).
Substituting \(R=r+6\): \((r+6)^2+r^2=116\), i.e. \(2r^2+12r+36=116\), so \(2r^2+12r-80=0\), i.e. \(r^2+6r-40=0\).
Solving: \(r=\dfrac{-6\pm\sqrt{36+160}}{2}=\dfrac{-6\pm14}{2}\); taking the positive root, \(r=4\).
So \(R=r+6=10\) cm.
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