Chain of Circles A chain of four circles centered at A, B, C, and D are touched on one side by the line GH and on the other side by a circular arc EF centered at O. Find the area of D in terms of the areas of A, B, and C
The hint is to find the relationship between the radii of those circles and to convert them to their areas. Would it have anything to do with the formula for their common external tangents $x = 2\sqrt{r_1 r_2}$? Anyone have an idea? If it is not so much to ask, I would appreciate it too if someone could give a detailed answer. I'm quite slow at following with geometry. :(
http://sphotos-h.ak.fbcdn.net/hphotos-ak-prn1/32333_108091916020416_1986142595_n.jpg
Here is what I have so far:
(1) (2sqrt(rd) + d – c)^2 + (r-d-2 sqrt(cd))^2 = (r+c)^2
(2) (2sqrt(rd) + d – b)^2 + (r-d-2sqrt(cd) – 2sqrt(bc))^2 = (r+b)^2
(3) (2sqrt(rd) + d – a)^2 + (r-d-2sqrt(cd)-2sqrt(bc)-2sqrt(ab))^2 = (r+a)^2
Would this get somewhere?
 A: Point 1: for two tangent circles, of radius $r_1$ and $r_2$, the distance between the points of tangency with a common external line is $2\sqrt{r_1r_2}$. This can be seen in this diagram:
$\hspace{4.5cm}$
Point 2: The centers of the circles are all on a parabola with focus $O$ and directrix $\overline{GH}$.
Point 3: For a quadratic function $f$ and three values $x$, $y$, and $z$,
$$
\frac{\dfrac{f(x)-f(y)}{x-y}-\dfrac{f(y)-f(z)}{y-z}}{x-z}\tag{1}
$$
is constant.
Thus, we have
$$
\frac{\dfrac{r_A-r_B}{\sqrt{r_Ar_B}}-\dfrac{r_B-r_C}{\sqrt{r_Br_C}}}{\sqrt{r_Ar_B}+\sqrt{r_Br_C}}
=\frac{\dfrac{r_B-r_C}{\sqrt{r_Br_C}}-\dfrac{r_C-r_D}{\sqrt{r_Cr_D}}}{\sqrt{r_Br_C}+\sqrt{r_Cr_D}}\tag{2}
$$
which can be solved as a quadratic in $\sqrt{r_D}$.
Simplification
The constant in $(1)$ is $\frac1{4f}$ where $f$ is the focal length of the parabola that passes through the centers of the circles. In fact, $(2)$ simplifies to
$$
\frac1f=\frac1{r_B}-\frac1{\sqrt{r_Ar_C}}=\frac1{r_C}-\frac1{\sqrt{r_Br_D}}\tag{3}
$$
Using $(3)$, it is easy to compute $f$ using $r_A$, $r_B$, and $r_C$, then it is easy to compute $r_D$ using $r_B$, $r_C$, and $f$.
Extension
With the bottoms of the circles on a horizontal line, circle $A$ on the left and $D$ on the right, the focus of the parabola is located at
$$
x=\frac{r_A-r_B}{\sqrt{r_Ar_B}}f+\sqrt{r_Ar_B}\qquad\text{and}\qquad y=f-\frac{x^2}{4f}\tag{4}
$$
relative to the center of circle $A$.
