Circle between curves exp(x) and ln(x) and lines.

Question

What is the radius of the circle inscribed between curves $\exp(x)$ and $\ln(x)$ and $x=5$ and $y=5$ lines.

Answer 1

The setup is symmetric with respect to the $x=y$ line, because $\exp$ and $\ln$ are inverse to one another. So the center will be at $(5-r,5-r)$. Furthermore, at the point of contact between circle and curve the tangent to the curve is perpendicular to the radius. So there is some point $(x,y)$ with $y=\exp(x)$ (point on the graph) and $(5-r-x)^2+(5-r-y)^2=r^2$ (right distance from center) and $(5-r-x)+(5-r-y)y=0$ (radius perpendicular to tangent direction $(1,y)$ i.e. slope $y$, expressed as zero dot product). This gives you three equations in three variables.

Is this explanation clear? Can you take it from here? I wouldn't be surprised if the transcendental nature of the exponential function leads to a transcendental solution with no easy formula for it, so you might want to concentrate on numeric approaches. I have computed that $$\small r=1.89205807597314114883620082321237421124261911223697300655781113454546\ldots$$ I could give you a lot more decimals, but no nice formula so far. To simply compute solutions you can use this bit of Python code:

from mpmath import mp, mpf, exp, findroot
def r(x):
    y = exp(x)
    return ((5-x)+(5-y)*y)/(1+y)
def f(x):
    y = exp(x)                         # first equation, as is
    r = ((5-x)+(5-y)*y)/(1+y)          # third equation, solved for r
    return (5-r-x)**2+(5-r-y)**2-r**2  # second equation, needs to be 0
mp.dps = 100  # number of digits, increase as you like
r(findroot(f, 1))

Exactly verifying that a given set of decimals is indeed correct requires some more work. I did that using interval arithmetic in Sage, so I can tell you that the digits I gave above are reliable.

Answer 2

Approaching this from a slightly different perspective, we could ask for the equation of the circle which is tangent to the graph of $y=e^x$ at some point $(x_0, y_0)$ and tangent to the graph of $y=\ln x$ at the point $(y_0, x_0)$. There is only one such circle, and we can easily find the center of this circle as it is the intersection point of the two normal lines through $(x_0, y_0)$ and $(y_0, x_0)$ respectively.

The center of the circle is then found to be $(h,h)$ where $$ h = \frac{y_0^2+x_0}{y_0+1}=\frac{e^{2x_0}+x_0}{e^{x_0}+1}$$

Having found the center of the circle it is trivial to determine the radius as its square is equal to $$ (h-x_0)^2 + (h-y_0)^2$$

With a bit of (computer) algebra we find

$$ r^2 = \frac{(y_0^2 + 1) (x_0 - y_0)^2}{(y_0 + 1)^2}=\frac{(e^{2x_0}+1)(x_0-e^{x_0})^2}{(e^{x_0}+1)^2}$$

In the problem posed by the OP we have $h+r=5$, so we're tasked with solving \begin{align*} \frac{e^{2x}+x}{e^{x}+1}+\sqrt{\left(\frac{(e^{2x}+1)(x-e^{x})^2}{(e^{x}+1)^2}\right)}&=5 \text{, or after some simplificaion}\\ \frac{e^{2x}+x+(e^x-x)\sqrt{e^{2x}+1}}{e^{x}+1} &=5\\ \end{align*}

A half-hearted attempt on the part of WolframAlpha finds $x\approx 1.28445$, and using this value to approximate $r$ gives $1.89205...$