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My Maths tutor showed me a shortcut way to find the equation of a tangent to a circle, given the radius, centre and point the tangent touches the circle:

$$(x - a)(c - a) + (y - b)(d - b) = r^2$$

where $(c,d)$ is the point the tangent touches the circle, and $(a,b)$ is the centre.

I've been searching for ages online for an example of this method, but every source I find uses the more "orthodox" method of finding the gradient of the radius and then using this to work out the gradient and then equation of the tangent.

Is it perfectly valid to use this equation in an exam, for example? And does it have a name?

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  • $\begingroup$ That is not an equation of a line. That's a circle. $\endgroup$ – GFauxPas Apr 19 '15 at 21:33
  • $\begingroup$ To find techniques related to this one (which as GFauxPas has noted is slightly misstated), a useful search term is "dot product". $\endgroup$ – user21467 Apr 19 '15 at 21:37
  • $\begingroup$ @StevenTaschuk What does the OP mean? $\endgroup$ – GFauxPas Apr 19 '15 at 21:42
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    $\begingroup$ As far as your second-to-last question is concerned, that seems better directed at your professor/teacher. :-) $\endgroup$ – Brian Tung Apr 19 '15 at 21:57
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    $\begingroup$ Very well, then I'll post an answer... $\endgroup$ – A.P. Apr 19 '15 at 22:06
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An elementary way to deduce this equation might be to observe that your circle has equation $$ r^2 = (x - a)^2 + (y - b)^2 = \color{red}{(x-a)}\color{green}{(x-a)} + \color{red}{(y - b)}\color{green}{(y - b)} $$ Since you want a line with exactly one point $(c,d)$ on the circle, it is reasonable to just evaluate the green part at $(c,d)$.


The more advanced answer is: actually, this is a direct application of the general formula for the tangent line of an algebraic curve.

For the purpose of this answer, let us define an algebraic curve as the zero locus (set) in $\Bbb{P^2(R)}$ of a homogeneous polynomial $p(X,Y,Z) \in \Bbb{R}[X,Y,Z]$ -- note that we can always get a curve in the affine plane by setting $Z = 1$, i.e. considering $p(X,Y,1)$. Then by Euler's theorem we know that $$ \frac{\partial p}{\partial X}(X,Y,Z) \cdot X + \frac{\partial p}{\partial Y}(X,Y,Z) \cdot Y + \frac{\partial p}{\partial Z}(X,Y,Z) \cdot Z = n p(X,Y,Z) $$ where $n$ is the degree of $p$. Thus we obtain the homogeneous equation of the line tangent to the curve $p(X,Y,Z) = 0$ at a point $(r,s,t)$ as $$ \frac{\partial p}{\partial X}(r,s,t) \cdot X + \frac{\partial p}{\partial Y}(r,s,t) \cdot Y + \frac{\partial p}{\partial Z}(r,s,t) \cdot Z = n p(r,s,t) = 0 \tag{1} \label{eq:tangent} $$ Now consider the polynomial $f(X,Y,Z) = (X - aZ)^2 + (Y - bZ)^2 - r^2 Z^2$, so that $f(X,Y,Z) = 0$ is the homogeneous equation of your circle. Then $$ \begin{align} \frac{\partial f}{\partial X} &= 2(X - aZ) \\ \frac{\partial f}{\partial Y} &= 2(Y - bZ) \\ \frac{\partial f}{\partial Z} &= -2a(X - aZ) - 2b(Y - bZ) - 2r^2 Z \\ \end{align} $$ Then we can use \eqref{eq:tangent} to find the equation of the affine tangent line at $(c,d)$: $$ \begin{align} 0 &= \frac{\partial f}{\partial X}(c,d,1) \cdot x + \frac{\partial f}{\partial Y}(c,d,1) \cdot y + \frac{\partial f}{\partial Z}(c,d,1) \cdot z \\ &= 2(c - a) x + 2(d - b) y - 2a(c - a) - 2b(d - b) - 2r^2 \\ &= 2\big( (c - a)(x - a) + (d - b)(y - b) - r^2 \big) \end{align} $$

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  • $\begingroup$ Incidentally, I think that equation \eqref{eq:tangent} is why Steven Taschuk mentioned the dot product: you can see it as the dot product between the gradient vector of $p$ and the "variables vector" $(X,Y,Z)$. $\endgroup$ – A.P. Apr 19 '15 at 22:41
  • $\begingroup$ If you find this answer hard to understand, please let me know. I'll try to make it more clear. $\endgroup$ – A.P. Apr 19 '15 at 22:59

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