Show that there are exactly two lines through a point p outside the circle that are tangent to the circle C Let $C$ be a circle of radius $r$ in the plane. Let $p$ be a point in the plane that lies outside of $C$. Show that there are exactly two lines through $p$ that are tangent to $C$.

It is one of those questions that seem very intuitive but very hard to prove for me. How do I show that there are "exactly" two tangent lines? Try to construct a third one but reach a contradiction? I'd appreciate any help. Thanks.

Method Using Calculus:  
Say we are given any circle and any point outside of that circle.  WLOG, we can translate the circle to be centered at the origin, and rotate our system so that the point is situated along the $y$-axis.  
From here, we claim that we can hit that point with exactly two tangent lines to the circle.  Imagine the circle is described by the equation $x^2 + y^2 = r^2$, and consider some point outside of the circle on the $y$-axis, say $(0, b)$.  
Differentiating implicitly, we find that $\frac{dy}{dx} = -x/y$.  Therefore, the tangent line through a given point on the circle, say $(m, n)$, will be described by the equation $y = (-m/n)x + b$, where $b$ is the $y$-intercept at the desired point.  Plugging in to solve for $b$:
$$n = -m^2/n + b$$
$$b = n+ m^2/n$$
We want to find points along our circle that both satisfy the above, and also $x^2 + y^2 = r^2$.  Rearranging the above, we get:
$$m^2 + n^2 = nb$$
In order to satisfy our circle's equation, we must take $n$ such that $nb = r^2 \Longrightarrow n = r^2/b$.  From here, we have exactly two choices for $m$ (positive or negative), and hence, exactly $2$ possible tangent lines to the circle through the desired point.
