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I have a question about some facts about curves. Consider a curve such that all the normals intersect at a fixed point. Prove that the trace of the curve it's contained in a circle. And prove that if a regular curve has the property that all the tangent lines pass through a fixed point, prove that the trace it's contained in a circle. Geometrically I can see it, especially the first and arguing with the fact that if the curve change it curvature then by the concavity or convexity then some normal vector does not intersect the same point of others. But exist some more formal proof?

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I assume by the curve's 'trace' you mean its image in Euclidean space.

(1) Let the fixed point be $\mathbf{p}$, and parametrize the curve by $\mathbf{x}(t)$. If the displacement vector going from $\mathbf{x}$ to $\mathbf{p}$ is parallel to the normal of the curve, then it must also be orthogonal to the tangent vector, and the tangent vector is parallel to $\mathbf{x}'(t)$. So we have

$$(\mathbf{x}(t)-\mathbf{p})\cdot\mathbf{x}'(t)=0 $$

The left hand side is actually the derivative of a simple expression. Thus, integrating yields

$$\frac{1}{2}\|\mathbf{x}(t)-\mathbf{p} \|^2=C.$$

This is precisely the equation of a circle of radius $\sqrt{2C}$ centered at $\mathbf{p}$.

(2) The curve here will actually be an open ray emanating outward from $\mathbf{p}$, if I'm understanding the question correctly. If the displacement $\mathbf{x}(t)-\mathbf{p}$ is parallel to the tangent vector, and so is $\mathbf{x}'(t)$, and the curve is regular, then there is a scalar function $\alpha(t)$ such that

$$\alpha(t)(\mathbf{x}(t)-\mathbf{p})=\mathbf{x}'(t).$$

This can be solved with the integrating factor method, but we'll take a shortcut. Reparametrize the curve so that the velocity is exactly $\mathbf{x}(s)-\mathbf{p}$, and then define $\mathbf{y}(s)=\mathbf{x}(s)-\mathbf{p}$. Then

$$\frac{d\mathbf{y}}{ds}=\mathbf{y}(s)$$

whose only solution is $\mathbf{y}(s)=e^s\mathbf{a}$ for some constant vector $\mathbf{a}$, which means that $\mathbf{x}(s)=\mathbf{p}+e^s\mathbf{a}$.

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