The problem:
Find the distance between the two identical cyan circles
I wonder if this problem could be done by the Descartes' circle theorem. This is my first time using this theorem and have just learned it from Brilliant. Anyway, I want to focus on this:
I want to treat the line $CD$ as an enormous circle so that the curvature is $0$. I just confuse determining which one I should put the negative sign. The article above says that since the circles are internally tangent to the biggest circle, I should put negative. Here's my attempt anyway:
$$\begin{align}
k &= -\frac 13 – \frac 19 \pm 2\sqrt{\frac 13\cdot \frac 19}\\
&= -\frac 49 + \frac 29\sqrt 3
\end{align}$$
Am I doing this right? Please don't bother solving this whole problem because I've already known the answer. I'm just looking for another alternative way by using Descartes' theorem if that's possible. The main problem here, is finding the radii of the cyan circle. If the DT can't be applied here, what's the simplest way to find the radii?
Best Answer
No you cannot apply Descartes' theorem for the circles with radius $9$, with radius $3$ and line $CD$ to find the radius of the cyan colored circle, as $CD$ is not tangent to the circle with radius $9$.
On your question of alternate approach to find radius $r$ of the cyan colored circle -
If $O, P , Q$ are respectively the centers of circles with radius $9, 3$ and $r$, note that
$PQ = 3 + r, PE = 3 - r$
Also, $OQ = 9 - r , OE = OC + CE = 3 + r$
Now applying Pythagoras theorem, $QE^2 = PQ^2 - PE^2 = OQ^2 - OE^2$
That gives you value of $r$. Once you know $r$, you can find the distance between identical circles which you mention you already know how to solve for.