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.

A proof using analytic geometry --
We can place the circle of radius $ \ r \ $ with its center at the origin, so its equation is $ \ x^2 \ + \ y^2 \ = \ r^2 \ $ .  We can also pick a point $ \ (C, 0 ) \ $ on the positive $ \ x-$ axis.  (This is equivalent to just saying we have some circle and some external point, so we'll define the center of the circle as the "origin" and the line through the center and the external point as the $ \ x-$ axis.  So far, this looks like Kaj_H's set-up.
A tangent line to the circle makes contact with it at a point $ \ (X,Y) \ $ , with the coordinates satisfying $ \ X^2 \ + \ Y^2 \ = \ r^2 \ $  .  The radius from the origin to that point has slope $ \ m \ = \ \frac{Y}{X} \ $ .   The  tangent line is perpendicular to that radius, so its slope is $ \ m' \ = \ \frac{1}{m} \ = \ -\frac{X}{Y} \ $ .  The equation of the tangent line is then
$$ y \ - \ 0 \ = \ \left(  -\frac{X}{Y} \ \right) \ \cdot \ ( \ x - C \ ) \ \ \Rightarrow \ \ y \ = \  \frac{XC}{Y}  \ - \ \frac{X}{Y} x  \ \ . $$ 
Now, a tangent point on that line is given by  
$$ Y \ = \  \frac{XC}{Y}  \ - \ \frac{X}{Y} \cdot X  \ \ \Rightarrow \ \ Y^2 \ = \ XC \ - \ X^2 \ \  \Rightarrow \ \ X^2 \ + \ Y^2 \ = \ XC \ \ . $$
[We know this manipulation is "safe" because a tangent line is not going to contact the circle at $ \ Y = 0 \ $ . ]
This means that $ \ XC \ = \ r^2 \  $ ; since we chose $ \ C \ $ to be positive, $ \ X \ $ must be as well.  But we also have that $ \ Y^2 \ = \ r^2 \ - \ X^2 \ $ .  A tangent line cannot contact the circle at $ \ X = 0 \ $ , as this would require a tangent line of slope zero (we see this from the line equation above).  Consequently, $ 0 \ < \ r^2 \ - \ X^2 \ < \ r^2 \ $ , so $ \ Y \ $ has two permissible values; thus, there are two possible tangent lines.
This argument is (unavoidably) related to those offered by ajotatxe and Kaj_H.
[A heuristic "proof":  I am standing outside a circular garden, loop of pavement, etc.  I see a left-hand edge and a right-hand edge.  Lines-of-sight to the edges follow tangent lines.]
