A problem about a tangent on $y=e^x$ and circles Let $\mathrm{A}$ be a point on the curve $y=e^x$. Then there are two circles $C_1$, $C_2$ such that tangents of them are the tangent of $y=e^x$ at the point $\mathrm{A}$ and $x$-axis, and they meet at $\mathrm{A}$. Assume that $C_1$ is bigger than $C_2$, WLOG.
Let $R$ be the radius of $C_1$. If $C_1$ and the curve $y=e^x$ meet at only two points --- it means that the number of intersections is $2$, prove that $R$ is the smallest radius which $C_1$ can have.
I found that $C_1$ and the curve $y=e^x$ can meet at only two points the case of 2 intersections by Intermediate value theorem. But I can't show why $R$ is the smallest. Could you please let me know why it holds?
(* Here is a graphical representation of this problem on GeoGebra.)
https://imgur.com/a/XfLBaKf
 A: Let $x$-axis is horizontal from left to right, and $y$-axis is vertical from bottom to top. Let's consider case when $C_1$ has exactly two common points with $y=e^x$. There is one common point in left half of circle $C_1$. So $A$ must be the only common point of $C_1$ and $y=e^x$ in right half of circle. $C_1$ and $y=e^x$ have common tangent in $A$, then $A$ is located in right bottom quarter of circle. Then $y=e^x$ doesn't have common point with top half of circle.
Let bottom half of $C_1$ is described by $y=y_1(x)$, top half of $C_1$ is described by $y=y_2(x)$ and center of $C_1$ is $(a,b)$. $C_1$ is tangent to $x$-axis, therefore radius of $C_1$ is $b$. Then the most right point of circle is $(a+b,b)$, $y_1(a+b)=y_2(a+b)=b$. Consider $f(x)=y_1(x)-e^x$ and $g(x)=y_2(x)-e^x$. $g(a)<0 \Rightarrow g(a+b)<0 \Rightarrow f(a+b) < 0$. $f(a) > 0$. Then $f(x)$ changes sign in point $A$. $y=y_1(x)$ and $y=e^x$ have common point in $A$ and have common tangent in $A$, then $f(x)=0$ and $f'(x)=0$ in point $A$. If $f''(x)\neq 0$ in $A$, then $A$ is point of extremum of $f(x)$, but this contradicts the fact that $f(x)$ changes sign in $A$. Therefore $f''(x)=0$ in $A$. Then in point $A$ must satisfy $y_1(x)=y_1'(x)=y_1''(x)=e^x$.
Let use parametrization of $C_1$: $y_1=b-b\cos t$, $x=a+b\sin t$. Then $y_1'=\frac{dy_1}{dx}=\tan t$, $y_1''=\frac{d(y_1')}{dx}=\frac1{b\cos^3 t}$. Then in point $A$ must satisfy $b-b\cos t=\tan t=\frac1{b\cos^3 t}$. Then $\cos t=\frac{2}{\sqrt{5}+1}$, $b=\sqrt{\frac{11+5\sqrt{5}}{2}}$. $a$ can be found from $\exp(a+b\sin t)=b-b\cos t$.
Let's consider case when $C_1$ has two or more common points with $y=e^x$ and one of them is $B$, where $C_1$ and $y=e^x$ have common tangent. Then in point $B$ must satisfy $y_1(x)=y_1'(x)=e^x$. Let use the same parametrization of $C_1$: $y_1=b-b\cos t$, $x=a+b\sin t$. Then in point $B$ must satisfy $b-b\cos t=\tan t$. Then $b=\frac{\tan t}{1-\cos t}$ and minimum $b$ can be found from $db/dt=0$, then $\cos t=\frac{2}{\sqrt{5}+1}$, $b_{min}=\sqrt{\frac{11+5\sqrt{5}}{2}}$.
