Find the equation of the osculating circle of the curve defined parametrically by :

$${displaystyle mathbf{r}(t),=,{begin{pmatrix}rleft(t-sinleft(tright)right)\rleft(1-cosleft(tright)right)end{pmatrix}}}$$

AKA cycloid.

The center $C$ of the osculating circle is given by:

$$C= mathbf{r}(t)+frac{1}{kappa } mathbf{N}(t)$$

Where $kappa $ is the curvature of the curve computed by :

$$kappa = frac{|mathbf{r}'(t)timesmathbf{r}”(t)|}{|mathbf{r}'(t)|^3}$$

And $mathbf{N}(t)$ is the unit normal vector defined by:

$$mathbf{N}(t) = frac{mathbf{r}'(t) times left(mathbf{r}”(t) times mathbf{r}'(t) right)}{left|mathbf{r}'(t)right| , left|mathbf{r}”(t) times mathbf{r}'(t)right|}$$

$${displaystyle mathbf{r}'(t),=,{begin{pmatrix}rleft(1-cosleft(tright)right)\rsinleft(tright)end{pmatrix}}};;;;,;;;;{displaystyle mathbf{r}”(t),=,{begin{pmatrix}rsinleft(tright)\rcosleft(tright)end{pmatrix}}}$$

$${displaystyle mathbf{r}'(t)timesmathbf{r}”(t) =,{begin{pmatrix}0\0\r^{2}left(cosleft(tright)-1right)end{pmatrix}}}$$

$$left|mathbf{r}'(t) right|

=rsqrt{2}sqrt{1-cosleft(tright)}$$

$$left|mathbf{r}'(t) times mathbf{r}”(t) right|

=r^{2}left|cosleft(tright)-1right|$$

$${displaystyle mathbf{r}'(t)times(mathbf{r}”(t) times mathbf{r}'(t))={begin{pmatrix}r^{3}left(sinleft(tright)-frac{sinleft(2tright)}{2}right)\r^{2}4sin^{4}left(frac{t}{2}right)\0end{pmatrix}}}$$

So the center of the circle is :

$$C= mathbf{r}(t)+frac{ left|mathbf{r}'(t) right|^3}{left|mathbf{r}'(t)times mathbf{r}”(t)right| }frac{ mathbf{r}'(t)times(mathbf{r}”(t) times mathbf{r}'(t)}{left|mathbf{r}'(t) right|left|mathbf{r}”(t)times mathbf{r}'(t)right| }$$

$$=mathbf{r}(t)+left(frac{ left|mathbf{r}'(t) right|}{left|mathbf{r}'(t)times mathbf{r}”(t)right| }right)^2mathbf{r}'(t)times(mathbf{r}”(t) times mathbf{r}'(t)$$

$$=,{begin{pmatrix}rleft(t-sinleft(tright)right)\rleft(1-cosleft(tright)right)end{pmatrix}}+left(frac{rsqrt{2}sqrt{1-cosleft(tright)}}{r^{2}left|cosleft(tright)-1right|}right)^{2}{begin{pmatrix}r^{3}left(sinleft(tright)-frac{sinleft(2tright)}{2}right)\r^{2}4sin^{4}left(frac{t}{2}right)end{pmatrix}}$$

The center simplifies to:

$$,{begin{pmatrix}rleft(t-sinleft(tright)right)\rleft(1-cosleft(tright)right)end{pmatrix}}+frac{csc^{2}left(frac{t}{2}right)}{r^{2}}{begin{pmatrix}r^{3}left(sinleft(tright)-frac{sinleft(2tright)}{2}right)\r^{2}4sin^{4}left(frac{t}{2}right)end{pmatrix}}$$

Let $frac{csc^{2}left(frac{t}{2}right)}{r^{2}}=v$.

Implies the equation of the circle:

$$left(x-color{blue}{left(r^{2}left(t-sinleft(tright)right)vleft(sinleft(tright)-frac{sinleft(2tright)}{2}right)right)}right)^{2}+left(y-color{blue}{left(r^{3}left(1-cosleft(tright)right)vleft(4sin^{4}left(frac{t}{2}right)right)right)}right)^{2}=left(frac{left(rsqrt{2}sqrt{1-cosleft(tright)}right)^{3}}{r^{2}left(1-cosleft(tright)right)}right)^{2}$$

I’m not sure if my work is right,can someone check that?