Let $\alpha:I\rightarrow\mathbb{R}^3$ be a regular curve. If there exists a point $P\in\mathbb{R}^3$ such that the line which passes through $P$ and $\alpha(t)$ is perpendicular to the tangent line at $\alpha$ in the instant $t$ for any $t$, then the trace of $\alpha$ is contained in a sphere.
As $(P-\alpha(t))\cdot \alpha'(t)=0$ and $((P-\alpha(t))\cdot (P-\alpha(t)))'=-2(P-\alpha(t))\cdot \alpha'(t)$, then
$$|P-\alpha(t)|^2=\int((P-\alpha(t))\cdot (P-\alpha(t)))'dt=\int-2(P-\alpha(t))\cdot \alpha'(t)dt=\int0dt=c$$ for some $c\geq0$.
Is it correct? How could I solve it without integrating?
You don't even have to know the concept of integration! Just use that fact that$$\left(\bigl\|P-\alpha(t)\right\|^2\bigr)'=0$$for each $t$, from which it follows that $t\mapsto\|P-\alpha(t)\bigr\|$ is constant.