If every tangent line to a curve $\Gamma$ runs through a common point ${\bf P}_o$, then $\Gamma$ is a line

Question

If every tangent line to a curve $\Gamma$ runs through a common point $\textbf{P}_o$, then that curve is a line.

Any help? I don't know how to start.

Answer 1

Let $P_0(x_0,y_0)$ and $y=y(x)$ the equation for the curve $\Gamma$.

$$y'(x)=\frac{y-y_0}{x-x_0}$$

$$\frac{dy}{y-y_0}=\frac{dx}{x-x_0}$$

$$\ln(y-y_0)=\ln(x-x_0)+k$$

$$y-y_0=C(x-x_0)$$

is the equation for a straight line. Being $C$ it's slope.

ADDED

I follow the differential idea because one of the tags the question has attached.

Answer 2

I am writing this answer but I may not have been able to frame it properly.

But,you can use my idea.

A tangent to a curve is given by the derivative of the curve's equation at that point.If $f(x)$ represents the curve and it has degree $>1$ then its derivative is never independent of $x$.But,if it a function $f(x)$ is linear (straight line) then it's derivative is a constant independent of $x$.

Hence,the result follows.