Three equal circles of unit radius touch one | vedclass

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(C) Let the centers of the three circles be $A, B,$ and $C$. Since each circle has a radius $r = 1$,the distance between any two centers is $AB = BC =

Vedclass · Accepted Answer

$\frac{\pi}{3}(2+\sqrt{3})^{2}$

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(C) Let the centers of the three circles be $A, B,$ and $C$. Since each circle has a radius $r = 1$,the distance between any two centers is $AB = BC = CA = 2r = 2$. This forms an equilateral triangle $ABC$ with side length $s = 2$. The centroid $O$ of this triangle is the center of the circumscribing circle. The distance from the centroid to any vertex (the circumradius $R_{tri}$ of the triangle) is given by $R_{tri} = \frac{s}{\sqrt{3}} = \frac{2}{\sqrt{3}}$. The radius $R$ of the large circumscribing circle is the sum of the distance from the centroid to a vertex and the radius of one of the small circles: $R = R_{tri} + r = \frac{2}{\sqrt{3}} + 1 = \frac{2+\sqrt{3}}{\sqrt{3}}$. The area of the circumscribing circle is $\pi R^2 = \pi \left( \frac{2+\sqrt{3}}{\sqrt{3}} \right)^2 = \pi \frac{(2+\sqrt{3})^2}{3} = \frac{\pi}{3}(2+\sqrt{3})^2$.

Three equal circles of unit radius touch one another. Then the area of the circle circumscribing the three circles is

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