<s>[INST] 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 
 [/INST] 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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