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Given three identical circles, with three points of intersection. The line between two of these intersecting points is $3$ feet. They are inside a $4$th circle. All circles are tangent to each other.

What is the area of the $4$th circle? I don't understand if any of the Descartes Theorem or Three Tangent Circles applies and how.

Or perhaps more simply, if the "outer Soddy circle" equation will yield the needed answer. works

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You can try getting the length from the centroid to a corner of the triangle formed by the centers of the three inner circles, and then add 3/2 to that to get the radius of the big circle. – J. M. May 2 '11 at 3:24
Can you see that the distance between the intersections equals the radius of the equal circles? – Gerry Myerson May 2 '11 at 4:18

Let's get a figure in here, eh?

tangent circles

We can exploit the fact that the points of mutual tangency of the three inner circles (red points in the figure) form an equilateral triangle; we also know that the red points are the midpoints of the segments formed by joining any two of the three blue points (the centers of the inner circles).

We then deduce that the triangle formed by the red points is half the scale of of the triangle formed by the blue points, and find that the equilateral triangle formed by the blue points has a side length of 6 ft., and that the inner circles have a radius of 3 ft. Using the law of cosines, we reckon that the distance from a blue point to the center of the triangle formed by the blue points is $2\sqrt{3}$ ft. Adding to that the radius of an inner circle, we find that the radius of the outer circle is $3+2\sqrt{3}$ ft.

The area of the outer circle is $\approx$ 131.27 square feet.

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M.: I think the OP meant that the distance between two of the points of tangency of the inner circles (with one another) is 3 ft. so the line you've drawn should be 6 feet, since the line the OP refers to has length = inner circles' radii.? Am I missing something? – amWhy May 2 '11 at 4:49
Hmm, I misread the problem I suppose; let me redo my computations... Thanks Amy! – J. M. May 2 '11 at 4:57

Given that the radii of the inner circles are $3$ feet, Descartes' Theorem (aka Soddy's Theorem) says that $$ \left(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}\right)^2=2\left(\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}+\frac{1}{r_4^2}\right) $$ Plugging in $r_1=r_2=r_3=3$, we get $$ \left(1+\frac{1}{r_4}\right)^2=2\left(\frac{1}{3}+\frac{1}{r_4^2}\right) $$ which turns into $$ r_4^2+6r_4-3=0 $$ which gives $r_4=-3\pm2\sqrt{3}$.

$r_4=-3+2\sqrt{3}$ says that the radius of the small inner circle tangent to the three given circles is $-3+2\sqrt{3}$.

$r_4=-3-2\sqrt{3}$ says that the radius of the outside circle is $3+2\sqrt{3}$. Thus, the area of the outside circle is $\pi(21+12\sqrt{3})$.

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