Fractal Orthogonal Circles

The exact answer is $R=\sqrt { \varphi +\sqrt { \varphi } } =1.70001577588....$ where $\varphi =\dfrac { 1 }{ 2 } \left( 1+\sqrt { 5 } \right)$ is the golden ratio.

Edit: Here's one method of computing $R$

Let $r$ be the radius of circle $n$ , and let $1$ be the distance between centers of circle $n$ and $n+1$ , and let $R$ be the ratio we seek. Thus, if circles $n$ and $n+1$ are orthogonal, we have

$r=\dfrac { 1 }{ \sqrt { 1+{ R }^{ 2 } } }$

Let $\theta$ be the angle between the lines from circle $n$ to $n+1$ and from circle $n+1$ to $n+2$ . Since circles $n$ and $n+2$ are also orthogonal, we have, using the Law of Cosines

$\theta =ArcCos\left( \dfrac { R }{ 1+{ R }^{ 2 } } \right)$

From this, and the property of self-similarity, we can compute the distance between the centers of circle $n$ and $n+4$ , which is

$\sqrt { { \left(\displaystyle \sum _{ k }^{ 4 }{ { R }^{ k-1 }Cos\left(k( \pi -\theta ) \right) } \right) }^{ 2 }+{ \left(\displaystyle \sum _{ k }^{ 4 }{ { R }^{ k-1 }Sin\left(k( \pi -\theta) \right) } \right) }^{ 2 } } =\sqrt { { \left( \dfrac { 1+{ R }^{ 4 } }{ 1+{ R }^{ 2 } } \right) }^{ 3 } }$

Since the radius of circle $n+4$ is ${ R }^{ 4 }r$ , and if circles $n$ and $n+4$ are tangent, then

$\dfrac { 1 }{ \sqrt { 1+{ R }^{ 2 } } } +\sqrt { { \left( \dfrac { 1+{ R }^{ 4 } }{ 1+{ R }^{ 2 } } \right) }^{ 3 } } =\dfrac { { R }^{ 4 } }{ \sqrt { 1+{ R }^{ 2 } } }$

for which the only real positive solution is $\sqrt { \varphi +\sqrt { \varphi } }$

The fact that the point where circles $n$ and $n+4$ are tangent happens to be the point where circles $n$ and $n+2$ intersect is a bonus and not necessary to find solution.

This is a five circles' question. After all, it seems to turn for about $15^\circ$ clockwise (to common convention), for every repeated enlargement. Named A, B, C, D and E from small to big, circle C coincides with circle A and circle E with $b^2 - 4 a c = 0$ confirms for a correct answer as a given clue. Set a good convention to solve this question numerically with geometry particularly for circle D and circle E, with Excel which can handle difficult figures:

1
2
3
4
5
6
7

A   0                   0                   1   
B   0                   -1.97232189012442   1.70001577588679    17000
C   3.01584883041592    -0.507016631010934  2.89005363826396    
D   -1.46530525911349   3.01584883041592    4.91313677820775    
E   -3.52286546142686   -6.45347597965383   8.35241003204278    
    -7.352410032        0.00000000000000E+00    
    0.479144314051276   0.877736137066449   2.89005363826396    Confirmed

What a golden ratio related answer! Checked that 2.89005363826396 to be correct and realized.

R = 1.700015775886789767192193615058+ and $\lfloor 10000 R\rfloor$ = 17000

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$Solving$ $resources:$

A (0, 0) of radius 1

B (0, - $\sqrt{1 + R^2}$ ) of radius R

C (R $\sqrt{1 + \frac{R^4}{1 + R^2}}, \frac{-1}{\sqrt{1 + R^2}}$ ) of radius R $^2$

We ought to first solve and visualize for center of circles A, B and C above.

With $x^2 + y^2 = 1$ centered at A (0, 0), B vertically below it would have a distance of $\sqrt{1^2 + R^2}$ , because tangent of circle A to center B ought to have a length of R which is the radius of circle B which is equals to the ratio to be determined. Since it forms a right angle triangle, distance A B and hence coordinates of B are determined as (0, - $\sqrt{1 + R^2}$ ). Circle B is $x^2 + (y + \sqrt{1 + R^2})^2 = R^2$ and we ought to point out center of circle C at C by drawing arcs to intercept but mathematically as:

$x^2 + y^2 = 1^2 + R^4$ and

$x^2 + (y + \sqrt{1 + R^2})^2$ = $R^2 + R^4$

$\implies (y + \sqrt{1 + R^2} - y)(y + \sqrt{1 + R^2} + y) = R^2 - 1$ by subtracting equation 2 with equaton1.

With working, y = $\displaystyle \frac{-1}{\sqrt{1 + R^2}}$ and hence x = $R \sqrt{1 + \frac{R^4}{1 + R^2}}.$ C (R $\sqrt{1 + \frac{R^4}{1 + R^2}}, \frac{-1}{\sqrt{1 + R^2}}$ ) of radius R $^2$ is found.

So far, $\pm$ of $\sqrt{?}$ are chosen according to diagram. For D and E to come, we should choose only $-\sqrt{?}$ for wanted cases. To intercept for D:

$(x - R \sqrt{1 + \frac{R^4}{1+ R^2}})^2 + (y + \frac{1}{\sqrt{1 + R^2}})^2$ = $R^4 + R^6$ and

$x^2 + (y + \sqrt{1 + R^2})^2 = R^2 + R^6$

Expand both equations and then subtract the first by the second. We shall obtain y = - x $\sqrt{R^2 + \frac{1}{R^2} + 1}$ .

Substitute y = - x $\sqrt{R^2 + \frac{1}{R^2} + 1}$ into expanded 2nd equation of $x^2 + y^2 + 2 \sqrt{1 + R^2} y + 1 = R^6$ , a quadratic equation is obtained:

$(R^2 + \frac{1}{R^2} + 2) x^2 - 2 \sqrt{(1 + R^2)^2 + 1 + \frac{1}{R^2}} x + 1 - R^6 = 0$ .

Let this complicated form be $a x^2 + b x + c = 0$ ,

Take negative, x = $\frac{- b - \sqrt{b^2 - 4 a c}}{2 a}$ and y = - x $\sqrt{R^2 + \frac{1}{R^2} + 1}$ . This is the numerical value of D which we can name it is D (-p, q).

Together with C (r, -s), of which p, q, r and s are taken as positive numbers to avoid ambiguity by choosing of signs as we have now known their values, arcs are drawn again to intercept for E:

$(x + p)^2 + (y - q)^2 = R^6 + R^8$ and

$(x - r)^2 + (y + s)^2 = R^4 + R^8$

Numerical instead of exact is becoming more demanding up to here as forms are complicated.

With y = u x + v = $\frac{p + r}{q + s} x + \frac12 \frac{p^2 + q^2 - r^2 - s^2 + R^4 - R^6}{q + s},$

$(1 + u^2) x^2 + 2 (u s + u v - r) x + r^2 + v^2 + 2 s v + s^2 - R^4 - R8 = 0.$

Again, we take negative, x = $\frac{- b - \sqrt{b^2 - 4 a c}}{2 a}$ and y = u x + v where u and v are dynamic values with respect to R being treated as constant, or a dynamic constant. In real access, value of p, q, r and s are $quoted from$ cell of -p, q, r, -s for easy references.

Numerical coordinates of $E$ in $dynamic$ form with respect to $R$ is obtained! To look for $b^2 - 4 a c = 0$ for one intercepting point between circle E and circle A:

$x^2 + y^2 = 1$ and

$(x - a)^2 + (y - b)^2 = (R^4)^2$

We can obtain a x + b y = $\displaystyle \frac{a^2 + b^2 + 1 - R^8}{2} = k$

Rearrange for y = $\frac{k - a x}{b},$

$x^2 + (\frac{k - a x}{b})^2 = 1$

$\implies (a^2 + b^2) x^2 - 2 k a x + k^2 - b^2 = 0.$

Set another $b^2 - 4 a c$ as a responsive reference for $R$ to look for a zero!

We shall find that only one excel value of $R$ = 1.70001577588679 can make a zero.

This solved for $R$ .

We shall also find that ( $\frac{k a}{a^2 + b^2}, \frac{k b}{a^2 + b^2})$ are coordinates of the only touch point which can also satisfy circle C; this $confirmed$ that $R$ = 1.70001577588679 is a correct finding. X (0.479144314051276, 0.877736137066449) for circle of radius = 2.89005363826396 suits.

Here ends. I found from author of this question that it is actually a golden ratio related value. His solution is an analytical result to solution like my solution here. I didn't apply rules for orthogonal interception for circles.

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Answer: $\boxed{17000}$