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Tangents to the circle

conscipost

Member
Jan 26, 2012
39
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

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

conscipost

Member
Jan 26, 2012
39
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,