A problem on four kissing circles (Descartes Theorem)

Two circles $C_1$ and $C_2$ with radius $r_1$ and $r_2$ touching each other externally and both touching circle $C_3$ with radius $r_1+r_2$ internally. If another circle with radius $r_3$ touches all three circles such that $r_1, r_2, r_3$ are in $A.P.$ and $r_1\gt r_2 \gt r_3$ find $$\frac{r_1}{r_2}$$

I tried applying the Descartes' theorem but the equation so formed seemed to be nearly unsolvable. Any new method or improvement in the method of Descartes theorem is appreciated.

• What is $A$? – Bernard Dec 20 '17 at 16:32
• It is not just A. It is A.P. i.e arithmetic progression – Rohan Shinde Dec 20 '17 at 16:34

Say $r_1 = k r_2$, so $r_3 = (2-k) r_2$. Observe first that the centres of $C_1$, $C_2$ and $C_3$ lie on a straight line, by considering the radii of each, so our diagram looks like this, where $D$ is the centre of $C_3$ From here, notice that $AD = r_1 + r_2 - r_3 = (2k-1)r_2$, and $CD= r_1 + r_2 - r_1 = r_2$. In particular, we know every length inside the triangle in terms of $r_2$and $k$. Apply Stewart's Theorem and cancel $r_2^3$ to give $(k+1)k + (2k-1)^2 (k+1) = 4k + (3-k)^2$, which simplifies to $k^3 = 2$, so $k = \sqrt[3]{2} = \frac{r_1}{r_2}$.