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

Title equation of tangent of circle EquationOfTangentOfCircle 2013-03-22 18:32:16 2013-03-22 18:32:16 pahio (2872) pahio (2872) 10 pahio (2872) Derivation msc 51M20 msc 51M04 Polarising