@Hippalectryon Firstly I have showed that $r(t)=(\cos^2 t-\frac{1}{2}, \sin t \cos t, \sin t)$
satisfies both of the given surfaces $x^2+y^2=\frac{1}{4}$ and $\left( x+\frac{1}{2}\right)^2+y^2+z^2=1$ and hence their intersection, that is $x^2+x=\frac{1}{2}$.
It remains to show that any point on that intersection is of the form $(\cos^2 t-\frac{1}{2}, \sin t \cos t, \sin t)$.
If $(x,y,z)$ lies on the sphere $\left( x+\frac{1}{2} \right)^2+y^2+z^2=1$ then $z^2=1-\left( x+\frac{1}{2} \right)^2-y^2 \leq 1$. Therefore $|z| \leq 1$. It follows that $z$ must be of the form $z= \sin t$ for s…