User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

= Tangent line to a circle = No idea whether this will work or not, but hey it's a sandbox, won't hurt to try

From a given point outside a given circle, it is possible to draw a tangent to that circle.

Proof: WLOG, let the circle in question be the unit circle centered at the origin. The equation of the circle is then

$x^2 + y^2 = 1$

Let the arbitrary point outside the circle be

$P_1 = (x_1,y_1)$

Case 1: The line in question is not vertical.

In that case, $x^2 + y^2 = 1$ represents an implicitly defined differentiable function of $x$. Taking the derivative WRT $x$ of both sides:

$2x + 2y \dfrac {\mathrm dy} {\mathrm dx} = 0$

$\dfrac {\mathrm dy} {\mathrm dx} = \dfrac {-x} {y}$

For any point $P_n = (x_n,y_n)$ on the circle other than $P = (1,0)$ and $P = (-1,0)$, the equation of the tangent line is

$y - y_n = \dfrac {-x_n}{y_n} (x - x_n)$, where $x_n^2 + y_n^2 = 1$

Now the union of all such lines is

$\R^2 \setminus \{(x,y) : x^2 + y^2 \le 1\}$

Which I don't know how to prove, and would probably be much more complicated than the geometric proof anyway. So this proof will soon be deleted, probably. $P_1$ was defined to be in this region, and so the line exists. --GFauxPas 11:27, 15 November 2011 (CST)