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The equation of a tangent line to a circle ( center at $(x_m, y_m)$ and radius $r$) at the point $(x_0,y_0)$ is given by:

\begin{align} (x-x_m)(x_0-x_m)+(y-y_m)(y_0-y_m)=r^2. \end{align}

How to derive this expression?

unfinished_sentenc
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  • Actually, the equation of a circle with center at $(x_m,y_m)$ and radius $r$ is $(x-x_m)^2+(y-y_m)^2=r^2$. – José Carlos Santos Aug 22 '17 at 14:36
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    You just did. You added two true equations to get a thrid true equation. – Somos Aug 22 '17 at 14:49
  • did you mean the missing + sign? – unfinished_sentenc Aug 23 '17 at 13:41
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    Don't delete a question after getting an answer. That gives a very bad aftertaste. As if you were trying to hide the fact that you asked this question. Also, you deprive the answerer a chance to get appreciation in terms of upvotes. Having most of your questions deleted also leads to a question ban, and your use of the site will be severely limited. – Jyrki Lahtonen Aug 24 '17 at 04:46

1 Answers1

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The equation of a circle with center $(x_m,y_m)$ and radius $r$ is given by $$ (x-x_m)^2+(y-y_m)^2=r^2\tag{1} $$ The derivative of $(1)$ with respect to $x$ is $$ \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x-x_m}{y-y_m}\tag{2} $$ where $(2)$ is derived by implicit differentiation of $(1)$ with respect to $x$. Hence, the gradient of the tangent line at $(x_0,y_0)$ is $$ m=-\frac{x_0-x_m}{y_0-y_m} $$ The equation of the tangent line at $(x_0,y_0)$ is then $$ y-y_0=-\frac{x_0-x_m}{y_0-y_m}(x-x_0)\Longleftrightarrow (x_0-x_m)(x-x_0)+(y_0-y_m)(y-y_0)=0\tag{3} $$ The point $(x_0,y_0)$ lies on the circle so it should satisfy $(1)$. $$ (x_0-x_m)^2+(y_0-y_m)^2=r^2\tag{4} $$ Add $r^2$ to both sides of $(3)$ to get $$ (x_0-x_m)(x-x_0)+(y_0-y_m)(y-y_0)+r^2=r^2\tag{5} $$ Using $(4)$, we can rewrite $(5)$ as $$ \begin{align} & \left((x_0-x_m)(x-x_0)+(x_0-x_m)^2\right)+\left( (y_0-y_m)(y-y_0)+(y_0-y_m)^2\right)=r^2\\ \Longleftrightarrow& (x_0-x_m)(x-x_0+x_0-x_m)+(y_0-y_m)(y-y_0+y_0-y_m)=r^2\\ \Longleftrightarrow& (x_0-x_m)(x-x_m)+(y_0-y_m)(y-y_m)=r^2 \end{align} $$ as required.