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Three circles of the same radius are arranged in such way that one circle is tangent to the other two. A fourth circle is drawn so that it will contain three circles and be tangent to the other three.If the smaller circles have a radius of 3cm. What is the area of the fourth circle?

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Given that you have three circles of the same radius, the tangency points form an equilateral triangle. For a more general case where the radii are not equal, you might find Descartes' theorem useful.

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@RodCarvalho: typo fixed. Thanks – Ross Millikan Nov 20 '12 at 1:05

The line extended through PO is the perpendicular bisector of the triangle. If we were to draw perpendicular bisectors through the other three vertices, we'd see that the perpendicular bisectors intersect at the origin. The origin is thus the centroid of the triangle, which means that it is 1/3 of the height of the triangle. Because the triangle is equilateral, we can see that its height is $3 \sqrt{3}$, so the centroid is $2\sqrt{3}$ cm away from the vertex B. B, in turn, is one radius, 3cm, away from the outer circle. Thus, the bigger circle has a radius of $3+2\sqrt{3}$.

It's area is simply $\pi r^2 = \pi(21+12\sqrt{3})$

enter image description here

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