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:


Differentiate (1), we have

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

At (x_0, y_0),

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


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


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