How to prove that only two tangents can be drawn from an external point to a circle?

Its obvious (by intuition) that from an external point(outside a circle) only two tangents can be drawn to the circle.

But , how do we prove it prove it?

N.B: 1. Please provide the simplest possible proof. 2. Here , it is proved for a parabola. But , how can we do the same for a circle.

Let $c$ be a circle whose center is $C$ and let $P$ be an external point to the circle. Let $T\in c$. Then the line defined by $P$ and by $T$ is tangent to $c$ if and only if that line is orthogonal to the line defined by $C$ and $T$. In other words, if and only if the angle $\angle C\hat TP$ is a right angle. And this will take place if and only if $T$ belongs to the circle $c^\ast$ of which the line segment $CP$ is a diameter. But $c\cap c^\ast$ consists of two and only two points.

Without the loss of generality say the circle is $x^2+y^2=r^2$. Let $P(a,b)$ be the external point. A line passing through $P$ with slope $m$ is of the form $$y-b=m(x-a).$$

To get the intersection points of this line and the circle, we substitute $y$ in the equation of the circle and obtain $$x^2(m^2+1)+2mx(b-am)+((am-b)^2-r^2)=0.$$

For tangency, we want equal roots. Thus we want the discriminant to be $0$. This gives us $$4m^2(b-am)^2-4(m^2+1)((am-b)^2-r^2)=0.$$ Upon simplifying we get, $$m^2r^2-(am-b)^2-r^2=0.$$ This is a quadratic equation in $m$ so at the most two real values. This means there can be only two possible tangent lines emanating from $P$.

In fact, it can be easily shown that if this equation has a real root then it will have two distinct real roots, thus exactly two tangents.