I want to calculate the radius of the smallest circle (radius $R$) that can hold $4$ circles (with radii $a, b, c, d$) inside it, such that:
No circles overlaps one other.
$a \ge b \ge c \ge d.$
Radius of $a, b, c, d$ are known to us.
We need to find $R$.
In this problem, the radius of $a,b,c,d$ are not always same.
I'm curious to find the solution for the following cases:
Case 1: $a = b = c = d$
Case 2: $a = b = c > d$
Case 3: $a = b > c \ge d$
Case 4: $a > b \ge c \ge d$
It is simple if $d$ is small enough. You can use Decartes' theorem to find the radius $r$ outside the $a,b,c$ circles as $(\frac 1a + \frac 1b + \frac 1c-\frac 1r)^2=2(\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+\frac 1{r^2})$ because the $d$ circle fits in one of the small spaces left over.