equation of tangent of circle

We derive the equation of tangent line for a circle with radius $r$.  For simplicity, we chose for the origin the centre of the circle, when the points  $(x,y)$  of the circle satisfy the equation

$x^2+y^2=r^2.$(1)

Let the point of tangency be  $(x_0,y_0)$.  Then the slope of radius with end point  $(x_0,y_0)$  is$\frac{y_0}{x_0}$, whence, according to the parent entry (http://planetmath.org/TangentOfCircle), its opposite inverse$-\frac{x_0}{y_0}$ is the slope of the tangent, being perpendicular to the radius.  Thus the equation of the tangent is written as

$$y-y_0=-\frac{x_0}{y_0}(x-x_0).$$

Removing the denominator and the parentheses we obtain from this first  $x_{0}x+y_{0}y=x_0^2+y_0^2$,  and then

$x_{0}x+y_{0}y=r^2$(2)

since  $(x_0,y_0)$  satisfies (1).

Remark.  In the equation (2) of the tangent, $x_0$, $y_0$ are the coordinates of the point of tangency and$x,y$ the coordinates of an arbitrary point of the tangent line.  But one can of course swap those meanings; then we interprete (2) such that $x_0$, $y_0$ are the coordinates of some point $P$ outside the circle (1) and $x,y$ the coordinates of the point of tangency of either of the tangents which may be drawn from $P$ to the circle.  If (2) now is again interpreted as an equation of a line (its degree (http://planetmath.org/AlgebraicEquation) is 1!), this line must pass through both the mentioned points of tangency $A$ and $B$ (they satisfy the equation!); in a , (2) is now the equation of the tangent chord $AB$ of  $P=(x_0,y_0)$.  See also http://mathworld.wolfram.com/Polar.htmlpolar.