Answer
The radius of curvature will become: $\rho=\dfrac{1}{\kappa}=1$ .
Center: $(\dfrac{\pi}{2}, 0)$.
So, our solution is: $(x-\dfrac{\pi}{2} )^2 +y^2=1$
Work Step by Step
$\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|$
where $T=\dfrac{v}{|v|} $, a unit tangent vector.
$v=\dfrac{dr}{dt}= i+ j \cos (t)$ and $|v|=\sqrt {(1)^2+(\cos t)^2}=\sqrt {1+\cos^2 t}$
So, $T=\dfrac{v}{|v|}=\dfrac{i+j \cos t }{\sqrt {1+\cos^2 t}}$
$\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|=(\dfrac{1}{\sqrt {1+\cos^2 t}} )( \dfrac{\sin t}{\cos^2 t +1})\\=\dfrac{\sin t \sqrt {\cos^2 t+1}}{(\cos^2 t+1)^2}$
Now, $\kappa (\dfrac{\pi}{2})=\dfrac{\sin (\dfrac{\pi}{2}) \sqrt {\cos^2 (\dfrac{\pi}{2})+1}}{[\cos^2 (\dfrac{\pi}{2})+1]^2}=1$
Thus, the radius of curvature will become:
$\rho=\dfrac{1}{\kappa}=1$ .
Center: $(\dfrac{\pi}{2}, 0)$.
So, our solution is: $(x-\dfrac{\pi}{2} )^2 +y^2=1$
