Area of a circle externally tangent to three mutually tangent circles 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 
 A: Let's get a figure in here, eh?

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.
A: Consider the three identical circles each of radius $r$ intersecting one another at three distinct points i.e. points of tangency. Join the centers of circles $A, B$ and $C$ to obtain the equilateral $\triangle ABC$. Join the points of tangency P and Q to obtain the equilateral $\triangle APQ$ whose each side is equal to the radius $r$ of each of three external tangent circles (as shown in the figure below).
$$r=PQ=3\mathrm{ft.}$$

In general, the radius $R$ of circle internally touching three mutually externally tangent circles of radii $a, b$ and $c$ is given by Generalized Formula
$$\boxed{R=\dfrac{abc}{2\sqrt{abc(a+b+c)}-ab-bc-ca}}$$
Now, substituting the value of radii identical circles i.e. $a=b=c=r=3$ in the above generalized formula as follows
$$R=\dfrac{3\cdot 3\cdot 3}{2\sqrt{3\cdot 3\cdot 3(3+3+3)}-3\cdot 3-3\cdot 3-3\cdot 3}=\dfrac{\sqrt3}{2-\sqrt3}=3+2\sqrt3$$
Therefore, the required area of the big circle with radius $R=3+2\sqrt3$ is
$$\pi R^2=\pi(3+2\sqrt3)^2=\pi(21+12\sqrt3)\ \mathrm{ft^2}$$
A: 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 $\pi(21+12\sqrt{3})$.
