# circles tangent to exponential curve

Circle $C_1$ is tangent to the curves $y=e^x$ and $y=-e^x$ and the line $x=0$, and for $n>1$ circle $C_n$ is tangent to both curves and to $C_{n-1}$, how can I find the radius of any circle $C_k$?

This is related to my previous question: summing series usings circles inside curves

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let $r$ be the radius of $C_1$ and $(x,e^x)$ the point of tangency. then we have $(r-x)^2+e^{2x}=r^2$ and $e^{2x}/(r-x)=1$ – yoyo Dec 26 '11 at 19:40

## 1 Answer

The circle with centre $(a,0)$ and radius $r$ is tangent to $x=0$ if $r=|a|$. It is then tangent to $y=e^x$ and $y=-e^x$ if there is $x$ such that $e^{2x} + (a-x)^2 = a^2$ and $2 e^{2x} + 2(x-a) = 0$. Eliminating $a$, the equation for $x$ is $(1-2x) e^{2x} - x^2 = 0$. This has two real solutions, approximately $-.7478435353725761$ and $.4580318389459811$, corresponding to $a = -.5237489523800181$ and $2.957464268427510$ respectively. I don't think there are "closed-form" solutions, even with the use of the Lambert W function.

Once you have the solution for the first circle, similar methods will get further circles. Thus the circle with centre at $(a_2, 0)$ and radius $r_2$ is tangent to $y=e^x$ and $y=-e^x$ and $x = 2 a$ (and thus to the first circle) if $r_2 = |a_2 - 2 a|$ and $e^{2x} + (a_2 - x)^2 = r_2^2$ and $2 e^{2x} + 2 (x - a_2) = 0$ For the case $a = -.5237489523800181$, I get $a_2 = -1.308841381001648$.

Here's a picture.

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pretty picture :) – mathmath8128 Dec 27 '11 at 1:22