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 $p_1>0$.
Best Answer
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 $OP$).