As shown in the diagram above, let $A$, $B$ and $C$ be the centers of the $3$ circles. Draw the lines $AC$ and $BC$ to join these centers. Since the circles are tangent to each other, these lines go through the points where they are tangent, so each length is the sum of the circle radii, i.e.,
$$|AC| = r_1 + r_3 \tag{1}\label{eq1B}$$
$$|BC| = r_2 + r_3 \tag{2}\label{eq2B}$$
Assign the point coordinates to be $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_3,y_3)$. With \eqref{eq1B} and \eqref{eq2B}, using these coordinates in the squares of the lengths of $|AC|$ and $|BC|$ gives
$$\begin{equation}\begin{aligned}
(x_1 - x_3)^2 + (y_1 - y_3)^2 & = (r_1 + r_3)^2 \\
x_1^2 - 2x_1x_3 + x_3^2 + y_1^2 - 2y_1y_3 + y_3^2 & = r_1^2 + 2r_1r_3 + r_3^2
\end{aligned}\end{equation}\tag{3}\label{eq3B}$$
$$\begin{equation}\begin{aligned}
(x_2 - x_3)^2 + (y_2 - y_3)^2 & = (r_2 + r_3)^2 \\
x_2^2 - 2x_2x_3 + x_3^2 + y_2^2 - 2y_2y_3 + y_3^2 & = r_2^2 + 2r_2r_3 + r_3^2
\end{aligned}\end{equation}\tag{4}\label{eq4B}$$
Next, \eqref{eq3B} minus \eqref{eq4B} gives
$$\begin{equation}\begin{aligned}
& x_1^2 - x_2^2 + 2(x_2 - x_1)x_3 + y_1^2 - y_2^2 + 2(y_2 - y_1)y_3 = r_1^2 - r_2^2 + 2(r_1 - r_2)r_3 \\
& 2(x_2 - x_1)x_3 + 2(y_2 - y_1)y_3 = r_1^2 - r_2^2 + 2(r_1 - r_2)r_3 - x_1^2 + x_2^2 - y_1^2 + y_2^2 \\
& c_1x_3 + c_2y_3 = c_3
\end{aligned}\end{equation}\tag{5}\label{eq5B}$$
where, to make the algebra easier to deal with, I have set
$$c_1 = 2(x_2 - x_1) \tag{6}\label{eq6B}$$
$$c_2 = 2(y_2 - y_1) \tag{7}\label{eq7B}$$
$$c_3 = r_1^2 - r_2^2 + 2(r_1 - r_2)r_3 - x_1^2 + x_2^2 - y_1^2 + y_2^2 \tag{8}\label{eq8B}$$
In \eqref{eq3B}, move the $x_1^2 + y_1^2$ terms to the right to get
$$-2x_1x_3 + x_3^2 - 2y_1y_3 + y_3^2 = c_4 \tag{9}\label{eq9B}$$
where
$$c_4 = r_1^2 + 2r_1r_3 + r_3^2 - x_1^2 - x_2^2 \tag{10}\label{eq10B}$$
Assuming $x_2 \neq x_1$, so $c_1 \neq 0$ in \eqref{eq6B}, you can get from \eqref{eq5B} that
$$x_3 = \frac{c_3 - c_2y_3}{c_1} \tag{11}\label{eq11B}$$
Substituting this into \eqref{eq9B} gives
$$\begin{equation}\begin{aligned}
& -2x_1\left(\frac{c_3 - c_2y_3}{c_1}\right) + \left(\frac{c_3 - c_2y_3}{c_1}\right)^2 - 2y_1y_3 + y_3^2 = c_4 \\
& -\frac{2x_1c_3}{c_1} + \left(\frac{2x_1c_2}{c_1}\right)y_3 + \frac{c_3^2}{c_1^2} - \left(\frac{2c_2c_3}{c_1^2}\right)y_3 + \left(\frac{c_2^2}{c_1^2}\right)y_3^2 - 2y_1y_3 + y_3^2 = c_4 \\
& \left(\frac{c_2^2}{c_1^2} + 1\right)y_3^2 + \left(\frac{2x_1c_2}{c_1} - \frac{2c_2c_3}{c_1^2} - 2y_1\right)y_3 - \frac{2x_1c_3}{c_1} + \frac{c_3^2}{c_1^2} - c_4 = 0 \\
& c_5y_3^2 + c_6y_3 + c_7 = 0
\end{aligned}\end{equation}\tag{12}\label{eq12B}$$
where, once again, to keep the algebra somewhat simpler, I've set
$$c_5 = \frac{c_2^2}{c_1^2} + 1 \tag{13}\label{eq13B}$$
$$c_6 = \frac{2x_1c_2}{c_1} - \frac{2c_2c_3}{c_1^2} - 2y_1 \tag{14}\label{eq14B}$$
$$c_7 = -\frac{2x_1c_3}{c_1} + \frac{c_3^2}{c_1^2} - c_4 \tag{15}\label{eq15B}$$
Using the quadratic formula in \eqref{eq12B} gives
$$y_3 = \frac{-c_6 \pm \sqrt{c_6^2 - 4c_5c_7}}{2c_5} \tag{16}\label{eq16B}$$
There are $2$ roots because circle $C$ could be located either below the tangent point between $A$ and $B$ or above it. Assuming you want the one below, choose the smaller root in \eqref{eq16B}. Finally, you can substitute that value into \eqref{eq11B} to get $x_3$. You will then have the coordinates of the center of the blue circle.
Let $R_g$ be the radius of the green circumcircle, $R_b$ be the radius of the single blue circle, and there be $n \ge 1$ red circles. We wish to solve $\theta_r$, the angle between centers of each pair of consecutive red circles, and $R_r$, the radius of each of the red circles.
The radius of the centers of the red circles must be $R_g - R_b$. If we use a coordinate system where the origin is at the center of the green circle, the center of the blue circle is at $(R_b-R_g, 0)$.
If $n = 1$, then the one red circle has radius $R_r = R_g - R_b$ and is centered at $(R_b, 0)$.
If $n = 2$, then the two red circles have radius $R_r$ centered at $(x, \pm y)$,
$$\left\lbrace\begin{aligned}
R_r &= \displaystyle \frac{4 R_g R_b (R_g - R_b)}{(R_g + R_b)^2 } \\
x &= \displaystyle \frac{R_g ( 3 R_b - R_g )}{R_g + R_b} \\
y &= R_r \\
\theta_r &= 2 \operatorname{atan2}(y, x) = 2 \arctan\left(\frac{4 R_b (R_g - R_b)}{(3 R_b - R_g)(R_b + R_g)}\right)\\
\end{aligned}\right.$$
For $n \ge 3$, we need to solve
$$\cos\left((n-1)\frac{\theta_r}{2}\right) - \frac{2 R_b}{R_g - R_b}\sin\left(\frac{\theta_r}{2}\right) + 1 = 0$$
for $\theta_r$; then $$R_r = \displaystyle R_g \frac{\sin\left(\frac{\theta_r}{2}\right)}{1 + \sin\left(\frac{\theta_r}{2}\right)}$$
I recommend using a binary search in range $$0 \le \theta_r \le \frac{2 \pi}{n}$$as there is only one zero for $$f(\theta_r) = \cos\left((n-1)\frac{\theta_r}{2}\right) - \frac{2 R_b}{R_g - R_b}\sin\left(\frac{\theta_r}{2}\right) + 1$$ If $f(\theta_r) \lt 0$, $\theta_r$ is too large; if $f(\theta_r) \gt 0$, $\theta_r$ is too small.
(The upper limit applies if $R_b = 0$, as then the red circles form a closed ring of circles.)
For neighboring red circles to touch, $$R_r = (R_g - R_r) \sin\left(\frac{\theta_r}{2}\right) \quad \iff \quad R_r = R_g \frac{\sin(\theta_r/2)}{1 + \sin(\theta_r/2)}$$
Here is a verified Python implementation, that returns $R_r$ and $\theta_r$ as a tuple:
from math import pi, atan2, sin, cos
def find_r_theta(rG, rB, n):
n = round(n)
if n < 1:
raise ValueError("N must be at least 1, %s given" % n)
if n == 1:
return (rG - rB, 0)
if n == 2:
rR = 4*rG*rB*(rG-rB)/(rG+rB)**2
return (rR, 2*atan2(rR, rG*(3*rB-rG)/(rG+rB)))
theta_max = 2*pi/n
theta_min = 0
ct = 0.5*(n-1)
cs = 2.0*rB / (rG - rB)
# 53 bits of precision
for k in range(0, 53):
theta = 0.5*(theta_min + theta_max)
d = cos(ct*theta) - cs*sin(0.5*theta) + 1
if d > 0.0:
theta_min = theta
elif d < 0.0:
theta_max = theta
else:
break
s = sin(0.5*theta)
rR = rG*s/(s+1)
# Abort if the blue radius is smaller than the red
if rB < rR:
raise ValueError("Blue radius (%f) smaller than the red radius (%d)" % (rB, rR))
return (rG*s/(s+1), theta)
If you add the following code, you can run it via python3 this.py rG rB N out.svg, specifying the green circle radius, blue circle radius, and the number of red circles, and it'll create an SVG image, out.svg, for illustration and verification; you can view those in any browser.
if __name__ == '__main__':
from sys import argv, stdout, stderr, exit
from math import ceil
if len(argv) < 4 or len(argv) > 5:
stderr.write("\n")
stderr.write("Usage: %s [ -h | --help ]\n" % argv[0])
stderr.write(" %s GREEN-RADIUS BLUE-RADIUS RED-CIRCLE-COUNT [ OUT.SVG ]\n" % argv[0])
stderr.write("\n")
exit(1)
rG = float(argv[1])
rB = float(argv[2])
n = round(float(argv[3]))
if len(argv) > 4:
svg = open(argv[4], mode="w", encoding="UTF-8")
else:
svg = stdout
rR, theta = find_r_theta(rG, rB, n)
center = int(ceil(rG+2))
svg.write('<?xml version="1.0" encoding="UTF-8" standalone="no"?>\n')
svg.write('<svg xmlns="http://www.w3.org/2000/svg" version="1.1" viewbox="0 0 %d %d">\n' % (2*center, 2*center))
svg.write('<rect x="0" y="0" width="%d" height="%d" fill="#ffffff" stroke="none" />\n' % (2*center, 2*center))
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#00cc00" />\n' % (center, center, rG))
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#0000ff" />\n' % (center-rG+rB, center, rB))
if (n & 1):
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#ff0000" />\n' % (center+rG-rR, center, rR))
for i in range(1, int((n+1)/2)):
a = i * theta
x = (rG - rR) * cos(a)
y = (rG - rR) * sin(a)
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#ff0000" />\n' % (center+x, center+y, rR))
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#ff0000" />\n' % (center+x, center-y, rR))
else:
for i in range(0, int(n/2)):
a = (i + 0.5) * theta
x = (rG - rR) * cos(a)
y = (rG - rR) * sin(a)
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#ff0000" />\n' % (center+x, center+y, rR))
svg.write('<circle cx="%.3f" cy="%.3f" r="%.3f" fill="none" stroke="#ff0000" />\n' % (center+x, center-y, rR))
svg.write('</svg>\n')
svg.flush()
if svg != stdout:
svg.close()
How did I find the solution for $n \ge 3$?
Consider the following illustration: The green circle is centered at origin, with radius $R_g$:
The blue circle has radius $R_b$ and is centered at $(R_b - R_g, 0)$. Note that its center $x$ coordinate is actually $c = -R_g + R_b$: blue radius in from the leftmost point on the green circle.
Angle $\theta$ forms an isosceles triangle, with sides $R_g - R_r$ and base $2 R_r$. If we split it into two right triangles, where the hypotenuse length is $R_g - R_r$, short side length is $R_r$, and the angle opposite the short side $\theta/2$. Thus,
$$\sin\left(\frac{\theta}{2}\right) = \frac{R_r}{R_g - R_r} \tag{1a}\label{G1a}$$
Solving this for $R_r$ yields
$$R_r = R_g \frac{\sin\left(\frac{\theta}{2}\right)}{1 + \sin\left(\frac{\theta}{2}\right)} \tag{1b}\label{G1b}$$
This means that the red circle radius $R_r$ is defined by $\theta$, and $\theta$ is our only free variable. And only positive $\theta$ make any sense.
If there was no blue circle at all, then the $n$ red circles would cover the full circle: $n \theta = 2 \pi$. This gives us the possible range for $\theta$,
$$0 \lt \theta \lt \frac{2 \pi}{n} \tag{2}\label{G2}$$
The topmost red circle is always at angle
$$\theta_T = \frac{(n - 1) \theta}{2}$$ counterclockwise from right. Its center is at $(x, y)$,
$$\left\lbrace \begin{aligned}
x = (R_g - R_r) \cos \theta_T &= (R_g - R_r) \cos\left(\frac{(n - 1) \theta}{2}\right) \\
y = (R_g - R_r) \sin \theta_T &= (R_g - R_r) \sin\left(\frac{(n - 1) \theta}{2}\right) \\
\end{aligned} \right. \tag{3a}\label{G3a}$$
For the topmost red circle and the blue circle to touch, we need the distance between their centers to match the sum of their radiuses. Squaring the distances, we have
$$(x - b)^2 + y^2 = (R_b + R_r)^2 \tag{3b}\label{G3b}$$
Next, we subtract the right side from the left side (so we get a function of form $f(\theta)$ whose root we need to find), substitute $x$, $y$, $b$, and $R_r$.
At this point, the expression starts to sprawl, and I for one switch to a Computer Algebra System. I suggest wxMaxima or SageMath; both free and available for all operating systems. In Maxima:
declare(R_g,real, R_b,real, R_r,real, theta,real, n,integer) $
R_r : R_g * sin(theta/2) / (1 + sin(theta/2)) $
x : (R_g - R_r) * cos((n-1)*theta/2) $
y : (R_g - R_r) * sin((n-1)*theta/2) $
b : -R_g + R_b $
EQ : (x-b)^2 + y^2 - (R_b + R_r)^2 = 0;
and after applying rational and trigonometric simplifications (trigsimp(ratsimp(EQ));), we get
(((2*R_g^2-2*R_b*R_g)*sin(theta/2)+2*R_g^2-2*R_b*R_g)*cos((n-1)*theta/2)
+(2*R_g^2-6*R_b*R_g)*sin(theta/2)
+4*R_b*R_g*cos(theta/2)^2+2*R_g^2-6*R_b*R_g) / (2*sin(theta/2) - cos(theta/2)^2 + 2) = 0
That divisor, $2\sin(\theta/2) - \cos(\theta/2)^2 + 2 \ge 1$ in this context, because we only do this for $n \ge 3$ and therefore $0 \lt \theta/2 \lt \pi/3$, per $\eqref{G2}$. So, we can just omit it, by multiplying the equation with it. There is also a common factor $R_g$, which we can divide out at the same time: trigsimp(ratsimp(% * (2*sin(theta/2) - cos(theta/2)^2 + 2) / R_g)); and we get
((2*R_g-2*R_b)*sin(theta/2)+2*R_g-2*R_b)*cos((n-1)*theta/2)
+ (2*R_g-6*R_b)*sin(theta/2)
+ 4*R_b*cos(theta/2)^2
+ 2*R_g-6*R_b = 0
Replacing $\cos(\theta)^2$ with $1 - \sin(\theta)^2$, we have
((2*R_g-2*R_b)*sin(theta/2)+2*R_g-2*R_b)*cos((n-1)*theta/2)
+ (2*R_g-6*R_b)*sin(theta/2)
+ 4*R_b*(1 - sin(theta/2)^2)
+ 2*R_g-6*R_b = 0;
and finally, if we ask Maxima to solve this for $\theta$, solve(%, theta);, it gives us two solutions:
[ theta = -%pi, cos((n-1)*theta/2) = -(2*R_b*sin(theta/2)-R_g+R_b)/(R_b-R_g) ]
The first one is garbage, so we grab the second one, cos((n-1)*theta/2) = -(2*R_b*sin(theta/2)-R_g+R_b)/(R_b-R_g);, and subtract right hand side from left hand side to get a nice function form: ratsimp(lhs(%) - rhs(%)) = 0;, so we get
((R_g-R_b)*cos((n-1)*theta/2) - 2*R_b*sin(theta/2) + R_g-R_b)/(R_g-R_b) = 0
i.e.
$$\frac{(R_g - R_b) \cos\left(\frac{(n-1) \theta}{2}\right) - 2 R_b \sin\left(\frac{\theta}{2}\right) + R_g - R_b }{R_g - R_b} = 0$$
or, equivalently
$$\cos\left(\frac{(n-1)\theta}{2}\right) - \frac{2 R_b}{R_g - R_b} \sin\left(\frac{\theta}{2}\right) + 1 = 0$$
Best Answer
A symmetry argument shows that the top two blue circles are tangent not just to the green circles on the same side of the chords, but also to the top green circle. This is because a reflection about a diameter that passes through, say, the center of the top right blue circle, must map that blue circle to itself, and the rightmost green circle to the topmost green circle, and vice versa.
Consequently, there is a horizontal tangent line that can be drawn so that the top five circles forms a third sector with all the corresponding tangencies preserved. This also implies that a circle centered at the origin that is externally tangent to all three green circles will have the same radius as the blue circle, since a reflection about a line joining the centers of the top green circle with the rightmost green circle will map the center of the top right blue circle to the origin.
Next, it is easy to determine the radius of the green cirles. Assuming without loss of generality that the large circle has unit radius. Then the radius of the green circle must satisfy
$$\frac{g}{2-g} = 1 - 2g, \implies g = \frac{3 - \sqrt{5}}{2}. \tag{1}$$
This in turn means that the blue circle has radius $$b = 1 - 2g = \sqrt{5} - 2. \tag{2}$$
The radius of the red circle $r$ is the most difficult to determine. One way is to place the figure on the coordinate plane and compute the coordinate of the center of the top right blue circle, which is $$(x_b, y_b) = \left(\sqrt{10 \sqrt{5} - 22}, 2(\sqrt{5} - 2) \right), \tag{3}$$ the proof of which I leave as an exercise (hint: compute the coordinate of the intersection of the horizontal chord with the chord on the right). Then $r$ must satisfy
$$\begin{align} x_r^2 &= (1-r)^2 - (b+r)^2, \\ (x_r - x_b)^2 + (b - r)^2 &= (b + r)^2, \end{align} \tag{3}$$ where $x_r$ is a nuisance variable representing the $x$-coordinate of the top right red circle. This gives us the unique nontrivial solution $$x_r = 4 \sqrt{\frac{\sqrt{5}-2}{5}}, \quad r = \frac{3 - \sqrt{5}}{10}. \tag{4}$$ hence $g/r = 5$.
For fun, I also made a little animation.