Circles on Circle!

As the situation is symmetrical so the center of each of the circle will lie on the Y axis.

Let the center of any general circle be $(0,C)$ and its radius be $r$

It has two points of intersection with the hyperbola which are again symmetrical so we can claim the points of intersection to be of the form $(\pm x , y)$ .

Our equation of a general circle of the sequence is:

$x^2+ (y-C)^2 = r^2$

Subtracting it with the equation of hyperbola we get:

$2y^2 -2Cy + C^2 -r^2+1 =0$

Now we can say that the above has two roots which are same. Thus the discriminant of the above quadratic equation is $0$ :

$4C^2 - 4\times 2 \times (C^2 - r^2 +1) = 0$

On evaluating we get the relation between the coordinates of center of circle and radius of circle:

$C = \sqrt{2r^2 -2}$

Also all the circles are tangent to each other so we can say that:

$C_n - C_{n-1} = r_n + r_{n-1}$

$\sqrt{2r_n ^2 -2} - \sqrt{2r_{n-1}^2 -2} = r_n + r_{n-1}$

As $n \to \infty$ we can say that $r_n \to \infty$

So, dividing the above relation by $r_{n-1}$ and let $\dfrac{r_n}{r_{n-1}} = x$ we get :

$\sqrt{2} x - \sqrt{2} = x +1$

which on evaluating gives

$x= \boxed{3 + 2 \sqrt{2}}$

If we define $u_n \,=\, 1 + 2r_1 + 2r_2 + \cdots 2r_n$ , then the circle $C_{n+1}$ will have centre on the $y$ -axis, and its point of tangency with $C_n$ will be at the point $(0,u_n)$ . Suppose that $C_{n+1}$ is tangent to the hyperbola at the points $(\pm\xi,\eta)$ .

The hyperbola has gradient $\tfrac{\xi}{\eta}$ at $(\xi,\eta)$ , and hence the normal there has equation $\eta x + \xi y \; = \; 2\xi\eta$ This normal meets the $y$ -axis at the point $(0,2\eta)$ , and this is the centre of $C_{n+1}$ . Thus $r_{n+1}^2 \; = \; \xi^2 + \eta^2 \; = \; 2\eta^2 + 1 \qquad \qquad r_{n+1} + u_n \; = \; 2\eta$ Thus $\begin{array}{rcl} 2\eta^2 + 1 & = & (2\eta - u_n)^2 \; = \; u_n^2 - 4u_n\eta + 4\eta^2 \\ 1+u_n^2 & = & 2(u_n - \eta)^2 \end{array}$ so that $\eta \,=\, u_n + \sqrt{\frac{1 + u_n^2}{2}}$ , and hence $u_{n+1} \; = \; u_n + 2r_{n+1} \; = \; 2(r_{n+1} + u_n) - u_n \; = \; 4\eta - u_n \; = \; 3u_n + 2\sqrt{2(u_n^2 + 1)}$ From this it is clear that $u_n \to \infty$ as $n \to \infty$ , and that $\lim_{n\to\infty}\frac{u_{n+1}}{u_n} \; = \; \alpha \; =\; 3 + 2\sqrt{2}$ Since $\frac{r_{n+1}}{r_n} \; =\; \frac{u_{n+1} - u_n}{u_n - u_{n-1}} \; = \; \frac{\frac{u_{n+1}}{u_n} - 1}{1 - \frac{u_{n-1}}{u_n}}$ we deduce that $\lim_{n \to \infty} \frac{r_{n+1}}{r_n} \; = \; \frac{\alpha - 1}{1 - \alpha^{-1}} \; = \; \alpha \; = \; \boxed{5.83}$ to $2$ DP.