How to prove that the line perpendicular to the radius is the tangent in the calculus sense? Let $P=(p_1,p_2)$ be a point on an semicircle and $r$ be the line perpendicular to the radius $\overline{OP}$, like the picture below.

Euclid showed (Book III, Proposition 16) that $r$ does not intersect the semicircle at any point other than $P$.
I'd like help to show that $r$ satisfies the calculus definition of the tangent line to the semicircle at $P$. For this, we have to show that if $y$ is the function whose graph is the semicircle, then
$$\lim_{x\to p_1}\frac{y(x)-y(p_1)}{x-p_1}$$
exists.
Since the slope of $\overline{OP}$ is $-p_2/p_1$ we know that the value of the limit have to be $\displaystyle \frac{p_1}{p_2}=\frac{p_1}{y(p_1)}$.
This is a part of the John Molokach's proof of the Pythagorean Theorem. So, we can not use the expression $y(x)^2=|\overline{OP}|^2-x^2$.
Thanks.

EDIT. Possible answer to my question: (Is it right?) It is enough to prove that the limits
$$\lim_{x\to p_1^+}\frac{y(x)-y(p_1)}{x-p_1}\tag{1}$$
and
$$\lim_{x\to p_1^-}\frac{y(x)-y(p_1)}{x-p_1}\tag{2}$$
exists and are equal.
Given $x\in(p_1,0)$, consider the lines $s(x)$, $t(x)$ and $u(x)$ as in the picture below.

Notice that
$$\text{slope of }t(x)\leq\text{slope of }s(x)\leq\text{slope of }r=\frac{p_1}{y(p_1)}.$$
So, if we assume $y$ continuous, we conclude that
$$\lim_{x\to p_1^+}[\text{slope of }t(x)]=\lim_{x\to p_1^+}\frac{x}{y(x)}=\frac{p_1}{y(p_1)}.$$
It follows from the squeeze theorem that
$$\lim_{x\to p_1^+}[\text{slope of }s(x)]=\frac{p_1}{y(p_1)}.$$
This shows that the limit $(1)$ exists and is equal to $p_1/y(p_1)$ because
$$\frac{y(x)-y(p_1)}{x-p_1}=\text{slope of }u(x)=\text{slope of }s(x).$$
The limit $(2)$ can be dealt analogously and a similar reasoning works if $p_1>0$.
 A: I don't think there is any calculus argument.  There is a classical geometry argument though.  
Suppose that the circle $\Gamma$ centered at $O$ with radius $OP$ intersects the line $r$ perpendicular to $OP$ at $P$ in another point $Q \neq P$.  Then, since the sum of all angles in the triangle $OPQ$ is $180^\circ$ and $OPQ$ is nondegenerate, we have $$\angle OQP = 180^\circ-\angle OPQ-\angle POQ =90^\circ-\angle POQ<90^\circ=\angle OPQ\,.$$
Since the greater angle of a triangle is subtended by a greater side, $OP<OQ$.  This contradicts the assumption that $Q$ is on $\Gamma$ (whence $OQ$ must equal $OP$).
A: I don’t know how much of the development of Calculus you want to use. If you’re willing to accept an answer that doesn’t proceed directly from the definition, but uses techniques at some remove from the definition, then try this:
You might as well assume that your circle is the unit circle, $x^2+y^2=1$. Using implicit differentiation, $2x\,dx+2y\,dy=0$, thus $\frac{dy}{dx}=-x/y$. The tangent to the circle at point $(a,b)$ thus has slope $-a/b$, while the slope of the radius is $b/a$. Each of these slopes is the negative reciprocal of the other, so the tangent is perpendicular to the radius.
A: Long after Euclid differential calculus came into being, it included differential generality about all curves and their curvature, not just circles.
Molokach looks at the tangent as 
1) Limiting case of the secant of coincident points is tangent ,
when
2) Radius of curvature is constant.
which is a typical view from differential calculus.
Let the slope at $r$ be a calculus definition of the tangent line to any curve at $P$. We have a normal to the tangent, there is no fixed radius or center..
$$ \lim_{x\to p_1}\frac{y(x)-y(p_1)}{x-p_1} = yd(x) $$
Curvature $ 1/\overline{OP}$ at P is ( s is arc length) is its change rate:
$$ \dfrac{d (tan^{-1} yd(x))  }  { ds } = \dfrac  { yd^{'} (x) }{  ( 1 + yd^2(x))^{3/2}}   $$
the circle is a special case when the above is a constant.
A: Here is a short outline of the proof that I think will answer your concern:


*

*Assume via Euclid that the tangent line to the semicircle is perpendicular to the radius drawn to the point of tangency.

*Slope of line containing radius is $\frac{dy}{dx}=\frac{y}{x}$, so slope of tangent line is $\frac{dy}{dx}=-\frac{x}{y}$.

*The general solution to the equation $\frac{dy}{dx}=-\frac{x}{y}$ is $x^2+y^2=C.$  Use the fact that the curve must pass through $(0,c)$ to solve for $C=c^2$ and then notice that the curve will pass through a point $(a,b)$ also so that $a^2+b^2=c^2.$


Essentially I created this argument with the notion that the PT can stem from the circle equation instead of the usual argument that uses the PT to write the equation of the circle.
As far as the exercise for the reader, that phrase was added by the editor of The American Mathematical Monthly as sort of a way to save some space in the page filler.  It would really only take a few lines to explain this...
First, if the point $P$ has coordinates $(x_1,y_1)$, then the equation of the tangent line is
$$y-y_1=\left.-\frac{x}{y}\right|_{(x_1,y_1)}(x-x_1).$$
Based on the previous argument, the line contains $P$ and its slope at $P$ is the same as the first derivative of the curve at $P.$
A: I fully agree with Batominovski that there is no calculus argument. For calculus argument to work one needs the functional relation between $x, y$.
It is easy to prove using Euclid axioms that if $OP$ is radius of a circle with center $O$ so that $P$ lies on the circle and $PT$ is any line through $P$ then line $PT$ intersects the circle at only one point $P$ if and only if $\angle OPT = \pi/2$. Such a line $PT$ is called tangent and based on figure if $P$ is $(x, y)$ then slope of $PT$ comes to be $-x/y = dy/dx$ and this gives $x^{2} + y^{2} = k$ and Pythagoras theorem can be established as mentioned in linked article in the question.
Update: We can further use the definition of tangent as given in calculus. Let $P(x, y)$ be a point on the curve and let $P'(x', y')$ be another nearby point on the curve. If the chord $PP'$ stabilizes in a unique direction as point $P'$ moves towards $P$ then the line through $P$ in that unique direction is said to be tangent to the curve at point $P$.
We add some rigor in the previous definition. If there is a unique line $PT$ passing through $P$ such that for any $\epsilon > 0$ there are points $P' P''$ on the curve with $P' \neq P \neq P''$ and $P$ on arc $P'P''$ such that the both the angles "between lines $PT$ and $PQ'$" and "between $PT$ and $PQ''$" where "$Q'$ is any point on arc $PP'$" and "$Q''$ is any point on arc $PP''$" are less than $\epsilon$ then we say that line $PT$ is a tangent to the curve at $P$.
Using this definition also we can show that tangent to a circle at a point is perpendicular to the radius passing through that point. Note further than this is still one step away from the concept of derivative. Here we are considering the existence of a unique direction in the limiting case when $P'$ moves towards $P$. Thus we are talking of the angle which the tangent makes with the $X$ axis.
In order to show that existence of tangent also leads to existence of derivative we need only need to worry about the special case when point $P$ lies on $X$ axis (i.e. when tangent is parallel to $Y$ axis) because $\tan \theta$ is undefined at $\theta = \pm(\pi/2)$. For other points on the circle $P$ the angle $\theta$ made by tangent at $P$ to the $X$ axis is such that $\tan \theta$ is defined and hence the slope/derivative $\dfrac{dy}{dx} = \tan \theta$ exists.
