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# 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, 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 fixed 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 is 1!), this line must pass through both the mentioned points of tangency $A$ and $B$ (they satisfy the equation!); in a word, (2) is now the equation of the tangent chord $AB$ of $P=(x_{0},\,y_{0})$. See also polar.

## Mathematics Subject Classification

51M20*no label found*51M04

*no label found*

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