Answer
$(x-\dfrac{\pi}{2} )^2 +y^2=1$
Work Step by Step
When $r(t)$ is a smooth curve, then the curvature can be defined as: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|...(1)$
where $T=\dfrac{v}{|v|} $, a unit tangent vector.
We have: $v=\dfrac{dr}{dt}= i+ j \cos (t)$ and $|v|=\sqrt {(1)^2+(\cos t)^2}=\sqrt {1+\cos^2 t}$
So, $T=\dfrac{v}{|v|}=\dfrac{i+j \cos t }{\sqrt {1+\cos^2 t}}$
Equation (1) becomes: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|=\dfrac{1}{\sqrt {1+\cos^2 t}} \times \dfrac{\sin t}{\cos^2 t +1}=\dfrac{\sin t \sqrt {cos^2 t+1}}{(\cos^2 t+1)^2}$
$\kappa (\dfrac{\pi}{2})=\dfrac{\sin (\dfrac{\pi}{2}) \sqrt {cos^2 (\dfrac{\pi}{2})+1}}{(\cos^2 (\dfrac{\pi}{2})+1)^2}=1$
Therefore, the radius of curvature is: $\rho=\dfrac{1}{\kappa}=1$ and the center is $(\dfrac{\pi}{2}, 0)$.
Now, the solution is: $(x-\dfrac{\pi}{2} )^2 +y^2=1$
