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

\displaystyle 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

\displaystyle 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 $A B$ of $P=(x_{0},\,y_{0})$ . See also http://mathworld.wolfram.com/Polar.htmlpolar.

Title	equation of tangent of circle
Canonical name	EquationOfTangentOfCircle
Date of creation	2013-03-22 18:32:16
Last modified on	2013-03-22 18:32:16
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Derivation
Classification	msc 51M20
Classification	msc 51M04
Related topic	Polarising