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 

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 $\pi(21+12\sqrt{3})$.
