Pushing the boundaries

Ok so let me post the solution try to understand without diagram

Solving for $C_1$

Let centre be (x,y) and radius $r$

As $C_1$ touches y-axis, x-coor. of centre = r

$C_1$ will touch both circles internally so

Distance between centres = difference in radius

$\displaystyle \sqrt{r^{2} + (y-2)^{2}} = 2 - r$

$\displaystyle \sqrt{r^{2} + (y+1)^{2}} = 3 - r$

Solving these 2 equations we get

$r = 12\sqrt{3} -20, y = 8 - 4\sqrt{3}$

r + 3y = 4

Solving for $C_2$

Let the centre be (X,Y) and radius 'R'

As $C_2$ touches x-axis, y-coor. = R

$C_2$ will touch the circle with centre(0,2) externally and circle with centre(0,-1) internally

$\displaystyle \sqrt{X^{2} + (R-2)^{2}} = 2 + R$

$\displaystyle \sqrt{X^{2} + (R+1)^{2}} = 3 - R$

Solving these 2 equations you will get

$R = \frac{1}{2}, X= 2$

Finally,

$\frac{X}{R} \times ( r + 3y) = \boxed{\boxed{16}}$

$C_{1}$ has center $((12\sqrt{3} - 20), (8 - 4\sqrt{3}))$ , and $C_{2}$ has center $(2, \frac{1}{2})$ .

Thus $(\dfrac{x_{2}}{y_{2}})*(x_{1} + 3y_{1}) = (\dfrac{2}{\frac{1}{2}})*(12\sqrt{3} - 20 + 24 - 12\sqrt{3}) = 4*4 = \boxed{16}$ .

I will post the solution method shortly, but for now, here is a diagram of the quartet of circles.

EDIT: I apologize for "pushing the boundaries" of everyone's patience in the refinement of this question. I appreciate all your input. :)