I have a problem involving spheres that can be simplified to that of two circles $A$ and $B$ with radii $a$ and $b$. You can imagine a cone (dashed lines) being defined between the extremities of the two circles along the line that passes through their centres.
I want to find and construct the minimum number of other circles along that line.
The conditions for the new circles are:
- Their radii interpolate between that of the radii of $A$ and $B$ along the connecting centre line
- Their extremities touch the sides of the 'cone edges'
- The centre of the new circles are positioned along the line, on the perimeter of the preceding circle
As the circles interpolate their radii I get:
radius = $a + (b - a)*\frac{i}{n}$
Where $n$ is the count of circles minus one (including $a$ and $b$), and $i$ is the integer index from $0$ to $n$ for the circle (including $a$ and $b$).
As I know the distance d between the centres of $A$ and $B$, if I consider the sum of the radii of the circles along the line that must equal $d$ I get a series:
$[a + (b - a)*\frac{i}{n}]*k^i + ... = d + b$
Where $k$ is the scaling factor that interpolates the radii of the circles between $A$ and $B$.
Now I need to find the minimum integer $n$, and along with that the associated value for $k$ (which is easy from one of the terms when $n$ is found). I am not sure where to go from here? I notice that series is almost geometric but each term also involves $i$a in the coefficient.
If as per your comments, you just want to fill the space with circles tangent to the borders, and to each other (except when no longer possible), then the calculation would be as such. For convenience, I am going to scale all distances by the radius $a$. Thus, the larger circle will always have radius $1$. For other scales, you will just have to multiply all distances by $a$.
By linearity, if a circle tangent to the two borders is placed with center on the medial line at a distance of $x$ from the center of $A$, then its radius should be given by $$r = 1 - \frac {x}d(1-b) = 1 - tx$$ where $t := \frac{1-b}d$. If the first circle placed is to be tangent to $A$, then the distance from its center to the center of $A$ is the sum of the two radii. That is, $x = r_1 + 1$. So $$r_1 = 1 - t(r_1 + 1)\\r_1(1+t) = 1-t\\ r_1 = \frac{1-t}{1+t}$$ It will be convenient to define $q = \frac{1-t}{1+t}$. So $r_1 = q$. Also $1 - q = \frac{2t}{1+t}$.
The second circle will be at a distance of $1+2q + r_2$, so $$r_2 = 1 - t(1+2q +r_2)\\r_2(1+t) = 1-t-2qt)\\r_2 = \frac{1-t}{1+t}-\frac{2qt}{1+t} = q\frac{1+t - 2t}{1+t} = q^2$$
If we assume that circle $i$ has radius $r_i = q^i$ for all $i< k$, then circle $k$ is at a distance of $$1 + r_k + 2\sum_{i=1}^{k-1} q^i =r_k - 1 + 2\sum_{i=0}^{k-1} q^i = r_k - 1 + 2\frac{1-q^k}{1 - q} = r_k + \frac{1 + q - 2q^k}{1-q}$$
So $$r_k = 1 - t\left(r_k + \frac{1 + q - 2q^k}{1-q}\right)\\ r_k(1+t)(1-q) = 1 - q - t - tq + 2tq^k = (1-t) - q(1+t) + 2tq^k\\ r_k(1-q) = \frac{1-t}{1+t} - q + \frac{2}{1+t}2q^k\\r_k(1-q) = q - q+ (1-q)q^k\\r_k = q^k$$
By induction, $r_k = q^k$ for all $k$. When circle $A$ is not scaled to a radius of $1$, this becomes $$q = \frac{d + b -a}{d+a-b}\\r_k = q^ka$$
Now the sum of the diameters of all the intermediary circles must be $\le d - a - b$ in order for the circles to fit. For $k$ intermediary circles,
$$2a\sum_{i=1}^k q^k \le d - a -b\\a2q\frac{1-q^k}{1-q} \le d - a -b$$ $\frac{2q}{1-q} = \frac{d - a + b}{a-b}$, so $$(1-q^k)a\frac{d - a + b}{a-b} \le d - a - b\\ 1 - \frac{(d - a - b)(a-b)}{(d-a + b)a} \le q^k$$ Taking the logarithm of both sides (and noting that since $q < 1, \log q < 0$), we get $$ k \le \dfrac{\log\left(1 - \frac{(d - a - b)(a-b)}{(d-a + b)a}\right)}{\log q}$$ Which gives the upper limit on how many intermediary circles can be added.
The total area of the intermediary circles (not including $A$ or $B$) will be $$\pi\sum_{i=1}^k (q^ia)^2 = \pi a^2q^2\frac{1-q^{2k}}{1-q^2}$$
Note that if you want a minimum on the area of the circles, you should exchange $a$ and $b$ in the above, as shifting the circles towards the smaller end, leaving the gap next to the large end circle, will make their total area go down.