Find the equation of the tangent at the point (x,y) on each of the curves: $x^2 + y^2 = c^2$ (Using the Limits Only!) Find the equation of the tangent at the point (x,y) on each of the curves:
$$x^2 + y^2 = c^2$$
This question is from The Book 'Differential Calculus for Beginners' by Joseph Edwards, given on page 13 of Chapter - 2.
This question was to be Evaluated using Limits (using Infinite Series) and not Derivatives.
Here's my try:
$$x^2 + y^2 = c^2 \Rightarrow y = \pm \sqrt{c^2 - x^2}$$
Let's Go only with the Positive ones for now:
$$\lim \limits_{h \rightarrow 0} [\frac{((1-(\frac{x+h}{c})^2)^\frac 1 2 - ((1-(\frac{x}{c})^2)^\frac 1 2 ))c}{h}]$$
Which I using the Binomial Expansion got this reduced to:
$$\lim \limits_{h \rightarrow 0} \frac l h(\frac{x^2 - h^2 - 2xh}{2c^2}+(...)\space \space )$$
Here the $(...)$ is some binomial expansion for the series (also l here is also some series which will is kept so as to be used in future if needed) which will be eliminated once the value of $h$ is just plugged.
So, I don't seem to get the solution equation which is: $Xx + Yy = c^2$
Is this the right direction? I dun think that this is a function, so I'm not sure that this method is correct either!
**Note: I don't think that the equation above has been made using that equation,but I don't know anything about it either. $x$ and $y$ are the points whose tangent we want to find **
 A: The courbes are circles centered at $(0,0),$ with radius $r.$
The tangent at a point of the circle is perpendicular to the corresponding diameter.
Thus $(x_0,y_0)$ are not only coordinates of a point of tangency, but also coordinates of the vector normal to the tangent at this point.
Hence the equation of the tangent at $(x_0,y_0)$ is $xx_0+yy_0=c.$ Uploading $x=x_0,y=y_0$ gives $c=r^2.$
The equation is $$xx_0+yy_0=r^2.$$
A: Denote $y(x) = \sqrt{c^2 - x^2}.\;$ The tangent at any point of the courbe for  $\;x\in (-c,c)$ exists and has the slope 
$$\lim \limits_{h \to 0} \left[\frac{((1-(\frac{x+h}{c})^2)^\frac 1 2 - ((1-(\frac{x}{c})^2)^\frac 1 2 ))c}{h}\right]$$ as written by OP. Compute this limit 
$$\begin{aligned}\lim_{h \to 0} \frac{\sqrt{c^2-(x+h)^2} - \sqrt{c^2-x^2}}{h}&=\lim_{h \to 0} \frac{c^2-(x+h)^2 - c^2+x^2}{h\left(\sqrt{c^2-(x+h)^2} + \sqrt{c^2-x^2}\right)}\\&=\lim_{h \to 0} \frac{-2x}{\sqrt{c^2-(x+h)^2} + \sqrt{c^2-x^2}}\\&=\frac{-x}{\sqrt{c^2-x^2}}=\frac{-x}{y}\end{aligned}$$
Can you finish from this?
At $(-c,0)$ and $(0,c)$ are tangents vertical, so their equations are $x=-c,\;x=c,\;$ respectively. These also can be rewritten in the form $Xx + Yy = c^2.$ 
For the half-circle under $x-$axis given by $y(x) = -\sqrt{c^2 - x^2}$ we obtain the equation analogously. 
