Can three tangents to a circle meet at a common point?

[poissonspot]

I considered the question of whether three tangents to the circle could meet at a common point and only came up with a contradiction by lengthy constructive means.
Circles are "nice", so there must be some clever ways of showing this fact that given a point outside the circle there are two tangent lines to the circle passing through this point. Any ideas? Thanks mucho,

 

[MarkFL]

You could, without loss of generality, use a unit circle, and a point on the $y$-axis outside of the circle $(0,b)$ where $1<b$.

The equation of the circle is:

\(\displaystyle x^2+y^2=1\)

The equation of the tangent lines is:

\(\displaystyle y=mx+b\)

Substitute for $y$ from the line into the circle:

\(\displaystyle x^2+(mx+b)^2=1\)

\(\displaystyle x^2+m^2x^2+2bmx+b^2=1\)

\(\displaystyle \left(m^2+1 \right)x^2+(2bm)x+\left(b^2-1 \right)=0\)

Now, by analyzing the discriminant, we find:

\(\displaystyle \Delta=(2bm)^2-4\left(m^2+1 \right)\left(b^2-1 \right)\)

\(\displaystyle \Delta=4\left(b^2m^2-b^2m^2+m^2-b^2+1 \right)\)

\(\displaystyle \Delta=4\left(m^2-b^2+1 \right)\)

We see that with $1<b$, there will be two real values of $m$, given by:

\(\displaystyle m=\pm\sqrt{b^2-1}\)

This implies there are two tangent lines:

\(\displaystyle y=\pm\sqrt{b^2-1}x+b\)

 

[poissonspot]

You could, without loss of generality, use a unit circle, and a point on the $y$-axis outside of the circle $(0,b)$ where $1<b$.

The equation of the circle is:

\(\displaystyle x^2+y^2=1\)

The equation of the tangent lines is:

\(\displaystyle y=mx+b\)

Substitute for $y$ from the line into the circle:

\(\displaystyle x^2+(mx+b)^2=1\)

\(\displaystyle x^2+m^2x^2+2bmx+b^2=1\)

\(\displaystyle \left(m^2+1 \right)x^2+(2bm)x+\left(b^2-1 \right)=0\)

Now, by analyzing the discriminant, we find:

\(\displaystyle \Delta=(2bm)^2-4\left(m^2+1 \right)\left(b^2-1 \right)\)

\(\displaystyle \Delta=4\left(b^2m^2-b^2m^2+m^2-b^2+1 \right)\)

\(\displaystyle \Delta=4\left(m^2-b^2+1 \right)\)

We see that with $1<b$, there will be two real values of $m$, given by:

\(\displaystyle m=\pm\sqrt{b^2-1}\)

This implies there are two tangent lines:

\(\displaystyle y=\pm\sqrt{b^2-1}x+b\)

Thanks, this is close to how I approached the problem. I used vector algebra though instead and solved for points where the dot product was equal to zero.

btw, I'm still interested in other solutions if anyone else is. Thx again,