Write the equation of a circle osculating with the curve f(x)=arctanx at the point x0=1.

The curve $f\left(x\right)=arc\tan x$.

We have to find the equation of an osculating circle at the point $x_0=1$.

Formula used:

Let $f\left(x\right)$ be any curve, the equation of an osculating circle at the point $\left(x_0, y_0\right)$ is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

Where,

$$\begin{gathered} h=x_0-\frac{1+{\left(y\text{'}\left(x_0\right)\right)}^2}{y\text{'}\text{'}\left(x_0\right)}.y\text{'}\left(x_0\right) \\ k=y_0+\frac{1+{\left(y\text{'}\left(x_0\right)\right)}^2}{y\text{'}\text{'}\left(x_0\right)} \\ r=\frac{1}{\left|\kappa \right|}, \kappa =\frac{y\text{'}\text{'}\left(x_0\right)}{{\left[1+{\left(y\text{'}\left(x_0\right)\right)}^2\right]}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}. \end{gathered}$$

$f\left(x\right)=arc\tan x$

The point $x_0=1$.

$\begin{array}{rcl}{y}_0& =& f\left(x_0\right)\\ & =& f\left(1\right)\\ & =& arc\tan \left(1\right)\\ y_0& =& \frac{\mathrm{\pi }}{4}\end{array}$

Differentiate $y=f\left(x\right)=arc\tan x$ with respect to $x$,

$y\text{'}=\frac{1}{1+x^2}$

$y\text{'}\left(1\right)=\frac{1}{1+1^2}=\frac{1}{2}$

Again differentiate $y\text{'}=\frac{1}{1+x^2}$ with respect to $x$,

$\begin{array}{rcl}y\text{'}\text{'}& =& -\frac{2x}{{\left(1+x^2\right)}^2}\\ y\text{'}\text{'}\left(1\right)& =& -\frac{1}{2}.\end{array}$