Tangent and Normal lines that pass through the origin 
Let $\alpha$ be a regular curve in $\mathbb{R}^2$ and let all of its tangent lines pass through the origin. Also, let $\beta$ be a regular curve in $\mathbb{R}^2$ and let all of its normal lines pass through the origin.

How can I show that $\alpha$ is contained in a straight line through
the origin and that $\beta$ is contained in a circle around the origin?
I know that the tangent line $\alpha(t)$ is the line that points in the direction of the tangent vector $T(t)$ and that the normal line of $\beta(t)$ is the line that points in the direction of $N(t)$, but how can I put this all together to complete the proof? 
 A: Let $c$ be a regular parametrization of the curve $\beta$. At each point of the curve a unique normal line passing through it is well defined as we are in $\mathbb{R}^2$ and there is a tangency direction because of the regularity of the parametrization. As the normal line at $c(t)$ passes through the origin it has the direction $c(t)-0 = c(t)$. And then, we deduce $\langle \dot{c}(t) , c(t) \rangle = 0$ because of them being orthogonal.
Applying this relation we get:
$\frac{d}{dt} \| c(t)\|^2 = \frac{d}{dt} \langle c(t) , c(t) \rangle = 2\langle \dot{c}(t) , c(t) \rangle = 0$
As the reasoning works for every $t$ we have that $\| c(t)\|^2$ is constant as a function of $t$ and so is $\| c(t)\|$. Putting $R$ for this last constant we have that $\beta$ is contained in the circumference of center $0$ and radius $R$.
A: The tangent line at the point $(t,\alpha(t))$ to the curve $\alpha$ passes through the origin and must have the equation
$$y=\alpha'(t)x$$
As the point $(t,\alpha(t))$ lies on this line the curve $\alpha$ satisfies
$$\alpha(t)=t\alpha'(t),~~~~~\forall t$$
Solve this to get the parametric equation of $\alpha$ in terms of $t$.
Similarly you can show that $\beta$ satisfies 
$$\beta(t)=\frac{t}{\beta'(t)}$$
and then solve this differential equation.
