# Constructing the tangent line of circle again

We will construct the tangent line of circle differently than what is shown in a previous post.

Start from the equation of circle with radius $r$, centered at the origin of the rectangular coordinate system:

$x^2+y^2=r^2\quad\quad\quad(1)$

Differentiate (1), we have

$2x + 2y { {dy} \over {dx} } = 0$.

At $(x_0, y_0)$,

${x_0 + y_0 {{dy} \over {dx}}} = 0$

i.e.,

$y_0 {{dy} \over {dx}} = -x_0$

If $y_0 \neq 0$,

${{dy} \over {dx} }={ {-x_0} \over {y_0}}$

$y = {{-x_0} \over {y_0}} x + m$

That is

$y_0 y = -x_0 x +my_0$

or

$x_0 x + y_0 y= my_0\quad\quad\quad(2)$

At $(x_0, y_0)$,

$my_0 = x_0^2+y_0^2=r^2$

It follows that by (2),

$x_0 x + y_0 y= r^2$