Show that all normals to $\gamma(t)=(\cos(t)+t\sin(t),\sin(t)-t\cos(t))$ are the same distance from the origin.

At the curve point $\gamma(t)$ we have the tangent vector $\dot\gamma(t)=(t\cos t,t\sin t)$. Turn this vector counterclockwise $90^\circ$, and obtain $(-t\sin t, t\cos t)$. When $t>0$ therefore the unit normal at $\gamma(t)$ is $n(t)=(-\sin t, \cos t)$. This allows to obtain the normal $\nu$ at $\gamma(t)$ in the parametric form $$\nu:\quad u\mapsto\nu(u)=\gamma(t)+u\,n(t)=\bigl(\cos t+(t-u)\sin t,\ \sin t-(t-u)\cos t\bigr)\ .$$ In order to determine the distance of $\nu$ from the origin $O$ we have to determine the point $P$ on $\nu$ for which $\vec{OP}\perp n(t)$. This means that we have to find the $u$-value for which $\nu(u)\perp n(t)$, or $$\nu(u)\cdot n(t)=-\sin t\bigl(\cos t+(t-u)\sin t\bigr)+\cos t\bigl(\sin t-(t-u)\cos t\bigr)=0\ .$$ This simplifies to $u=t$, so that we obtain $P=\nu(t)=(\cos t,\sin t)$. This shows that $|OP|=1$, independently of $t>0$. The case $t<0$ is of course analogue. (At $\gamma(0)$ the curve has a singularity.)