# Find the radius of the n+1th circle if it is constructed to be internally tangent to circle number 1, externally to circle number n and number 2.

Given a circle with radius $$r_1$$ a circle (number 2) with radius $$r_2 < r_1$$ is drawn such that it is tangent to circle internally. a circle (number 3) is constructed with radius $$r_3= r_1 - r_2$$ such that it is tangent to first circle internally and the second circle externally .

Now the question is find the radius of the $$n$$th circle where the fourth circle tangent to circle number 1 internally and tangent to circle number 2 and number 3 externally.

the $$n+1$$th circle is constructed to be tangent to circle number 1 internally and tangent to circle number $$n$$ and number $$2$$ externally.

Finding $$r_4$$ was very hard. Solving this problem without coordinates will be a nightmare I tried to find the radius of the fourth circle by solving for $$r_4$$ in these three equations

$$y^2 +(x-r_1)^2 = (r_1-r_4)^2$$ $$y^2 +(x-r_2)^2 = (r_2+r_4)^2$$ $$y^2 +(x+r_2-r_1)^2 = (r_1-r_2+ r_4)^2$$

But I wasn't able to find a closed form for $$r_4$$.

• I feel your third circle must have a constraint (otherwise $C_3$ could be $C_5$, for example). A "natural" restriction could be the collinearity of centers $C_1,C_2,C_3$ (circle $C_3$ of largest area possible). Commented Apr 26 at 21:31
• Other necessary restriction is that radii $r_2\gt r_3$ Commented Apr 26 at 21:38
• @Piquito I forgot to write that
– pie
Commented Apr 27 at 8:52
• en.m.wikipedia.org/wiki/Pappus_chain
– ACB
Commented Apr 27 at 13:22
• You mean the link? I just googled 'Pappus chain' :) I had previously come across some stuff related to this.
– ACB
Commented Apr 27 at 15:32

If all you want are the radii, then Descartes' circle theorem will suffice. Let $$k_i = \pm 1/r_i$$ be called the curvature, where $$r_i$$ is the radius of circle $$i$$. The sign of $$k_i$$ depends on whether the tangency is internal or external to the respective circle. So in your case, circle $$C_1$$ has radius $$r_1$$ but the curvature should be $$k_1 = -1/r_1$$, because all of the other circles are internally tangent to $$C_1$$. Since all other tangencies are external, all the other $$k_i$$ except $$k_1$$ will use the positive curvature.

Then Descartes' theorem says $$(k_1 + k_2 + k_3 + k_4)^2 = 2(k_1^2 + k_2^2 + k_3^2 + k_4^2), \tag{1}$$ for four mutually tangent circles $$C_1, C_2, C_3, C_4$$. Note that in your case, you have assumed that the centers $$C_1, C_2, C_3$$ are collinear, which is not explicitly stated in your problem. In such a case, then $$(1)$$ may be rewritten in terms of the radii $$r_1$$ and $$r_2$$ as: $$\left(-\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_1 - r_2} + \frac{1}{r_4}\right)^2 = 2\left(\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{(r_1 - r_2)^2} + \frac{1}{r_4^2}\right). \tag{2}$$ Solving this equation for $$r_4$$ leads to two solutions, one positive, one negative. The positive root is the radius of $$C_4$$ with external tangency. The solution may be further simplified by assuming without loss of generality that $$r_1 = 1$$.

In this manner, we can recursively solve for $$r_n$$ in terms of $$r_1$$, $$r_2$$, and $$n$$, and derive a general formula. The details of this computation are left as an exercise.

In case you want the coordinates of the centers of the $$C_i$$, then the Wikipedia page also describes the complex-valued generalization to Descartes' theorem, which allows us to obtain the centers through a two-step process when placing the circles in the complex plane: first, find the radius using the regular theorem, then use the generalization to obtain the coordinates of the next center. Again, the details are left as an exercise.

As an additional note, the chain of circles you have described are inscribed in shape called an arbelos, and that circle tangencies of this sort are known as Apollonian packings or Soddy circles. A solution for this particular chain can also be obtained via inversive geometry through a suitably chosen circle of inversion.

Let the four circles $$\Gamma_1:(X-a)^2+Y^2=a^2\hspace 1cm\Gamma_2:(X-b)^2+Y^2=(2a-b)^2\\\Gamma_{n-1}:(X-c)^2+(Y-d)^2=k^2\hspace1cm \Gamma_n:(X-x)^2+(Y-y)^2=z^2$$

The actual calculation of the center and radius of the $$n^{th}$$ circle is very ugly. Its three equations are easy to set up since the center must be equidistant from the circles $$\Gamma_1,\Gamma_2$$ and $$\Gamma_{n-1}$$. One has $$\begin{cases}(x-a)^2+y^2=(a-z)^2\\(x-b)^2+y^2=(2a-b+z)^2\\(x-c)^2+(y-d)^2=(k+z)^2\end{cases}$$ from which you have $$\begin{cases}z=\dfrac{(b-a)(x-2a)}{b-3a}\\y=\left(\dfrac{(3c+k)a-(c+k)b-2a^2}{d(b-3a)}\right)x+\dfrac{2kba+2(b-k)a^2-2a^3}{d(b-3a)}+\dfrac{c^2+d^2-k^2}{d}\\x=\sqrt{(a-z)^2-y^2}+a\end{cases}$$ Whoever wants (not me) can achieve a resultant equation in $$x$$ with the way everybody knows and show the expressions of $$x,y,z$$ as functions of $$a,b,c,d,k$$

Another story is to try this problem with numbers instead of algebraic letters. For example, with $$a=5,b=7$$ you get $$\Gamma_3: \left(X-\dfrac{70}{19}\right)^2+\left(Y-\dfrac{60}{19}\right)^2=\left(\dfrac{30}{19}\right)^2$$ and you can try now to find out the corresponding circle $$\Gamma_4$$