Tangent graph to a circle

Question

I am trying to find the positive real number $b$ such that the graph of $f(x)=\ln(x+b)$ is tangent to the circle $x^2+y^2=1$. Here is my attempt:

I realize that the point of tangency must be in the top half of the circle if $b$ is positive, so I define $g(x)=\sqrt{1-x^2}$ (top half of circle). Now, I THINK that if two graphs $f$ and $g$ are tangent to each other at $x=a$, then $f(a)=g(a)$ and $f'(a)=g'(a)$ (correct me if I am wrong). Now, $f'(x)=\frac{1}{x+b}$, and $g'(x)=\frac{-x}{\sqrt{1-x^2}}$. So, we are basically solving for the $x$ value of the point of tangency and $b$. Now, I am not sure where to go from here. I used a graph to estimate a solution between $b=2.9$ and $b=3$. Can someone help?

score 2 · Accepted Answer · answered May 26 '19 at 15:18

2

Solving the equation $f'(x)=g'(x)$ for $b$, you get $$ b = -x - \frac{\sqrt{1-x^2}}{x} $$ Plug that in to $f(x) = g(x)$ and you have $$ \ln \left(-\frac{\sqrt{1-x^2}}{x}\right) = \sqrt{1-x^2} $$ This is unlikely to have closed-form solutions, but you can use numerical methods, such as Newton's. Maple tells me $$ x = -0.3669416757$$ which corresponds to $$ b = 2.902069222$$

answered May 26 '19 at 15:18

Robert Israel

448,999

Wait, does that mean that my intuition was right? – Jithinash May 26 '19 at 16:10
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That $f(a)=g(a)$ and $f'(a)=g'(a)$ means the curves $y=f(x)$ and $y=g(x)$ are tangent at $x=a$? Yes. – Robert Israel May 27 '19 at 01:43

score 0 · Answer 2 · answered May 27 '19 at 06:26

Satrting from Robert Israel's answer, you look for the zero of function $$f(x)=\ln \left(-\frac{\sqrt{1-x^2}}{x}\right) - \sqrt{1-x^2}$$ Plotting the function, we can see that the root is quite close to $-0.4=\frac 25$.

Peform a Taylor expansion around the value to get $$f(x)=\left(\log \left(\frac{\sqrt{21}}{2}\right)-\frac{\sqrt{21}}{5}\right)+\left(\frac{12 5}{42}-\frac{2}{\sqrt{21}}\right) \left(x+\frac{2}{5}\right)+\left(\frac{8125}{3528}+\frac{125}{42 \sqrt{21}}\right) \left(x+\frac{2}{5}\right)^2+O\left(\left(x+\frac{2}{5}\right)^3\right)$$ which is a quadratic in $\left(x+\frac{2}{5}\right)$.

Solve it to get $x\approx -0.366863\implies b\approx 2.90262$ which is not too bad.

For sure, you can polish the root using Newton method and using the guess for $x_0$, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & -0.36686326831355560408 \\ 1 & -0.36694166770061389757 \\ 2 & -0.36694167568391188355 \\ 3 & -0.36694167568391196631 \end{array} \right)$$

Tangent graph to a circle

2 Answers2