An argument for a line being tangent to a circle in plane geometry Here is a proposition presented in every plane geometry textbook.  The ones I saw containing a demonstration for it were not appealing to me.  I would appreciate any helpful comments to the demonstration that I am providing.
$P$ is a point on a circle centered at O, and $\ell$ is a line containing $P$. $\ell$ is tangent to the circle at $P$ if, and only if, the line is perpendicular to radius $\overline{OP}$.
Demonstration of ``If''
$\ell$ is tangent to the circle at $P$, $m$ is the line through $P$ that is perpendicular to $\ell$, and $n$ is the line through $O$ that is perpendicular to $m$. Two lines that are perpendicular to a third line are parallel to each other. So, $\ell$ is parallel to $n$.
If $m$ were not to contain $O$, the intersection $Q$ of $m$ and $n$ would be distinct from $O$. The distance $d$ between $Q$ and $\ell$ is the length of the line segment $\overline{PQ}$. According to the Pythagorean Theorem,
\begin{equation*}
\bigl\vert \overline{OP} \bigr\vert^{2} = \bigl\vert \overline{OQ} \bigr\vert^{2} + \bigl\vert \overline{PQ} \bigr\vert^{2}
< \bigl\vert \overline{PQ} \bigr\vert^{2} .
\end{equation*}
(The inequality here is a ridiculous mistake!)  So, $\overline{OP}$ is a shorter line segment than $\overline{PQ}$. The distance $d^{\prime}$ between $O$ and $\ell$ is less than or equal to the distance between $O$ and $P$, a particular point on $\ell$. So,
\begin{equation*}
d^{\prime} \leq \bigl\vert \overline{OP} \bigr\vert < \bigl\vert \overline{PQ} \bigr\vert = d .
\end{equation*}
Points on one of two parallel lines are a common distance from the other line. Though, $O$ is closer to $\ell$ than $Q$. This is a contradiction. So, $O$ is a point on $m$, and the radius $\overline{OP}$ is perpendicular to $\ell$. 
Demonstration of ``Only If''
$\ell$ is perpendicular to radius $\overline{OP}$, and $Q$ is any point on $\ell$ distinct from $P$. According to the Pythagorean Theorem,
\begin{equation*}
\bigl\vert \overline{OQ} \bigr\vert^{2} = \bigl\vert \overline{PQ} \bigr\vert^{2} + \bigl\vert \overline{OP} \bigr\vert^{2}
> \bigl\vert \overline{OP} \bigr\vert^{2} .
\end{equation*}
So, $\bigl\vert \overline{OQ} \bigr\vert > \bigl\vert \overline{OP} \bigr\vert$, and $Q$ is outside the circle. The line and circle intersect only at $P$, and $\ell$ is tangent to the circle.
The following code will be rendered by TikZ to give a diagram associated with this argument.
\noindent \hspace*{\fill}
\begin{tikzpicture}
\coordinate (O) at (0,0);
\draw[fill] (O) circle (1.5pt);
\draw[name path=circle] (O) circle (2);
%Parallel lines k and $\ell$ are drawn. A and B are the intersections of k and the circle, and
%D and E are intersections of $\ell$ and the circle.
\coordinate (A) at (180:2);
\coordinate (B) at (70:2);
\draw (A) -- (B);
\coordinate (C) at (235:2);
\path[name path=path_to_locate_D] let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)} in (C) -- ($(C) +(\n1:4)$);
\coordinate[name intersections={of=path_to_locate_D and circle}];
\coordinate (D) at (intersection-1);
\draw (C) -- (D);
%The labels for A, B, C, D, and O are typeset.
\coordinate (label_for_A_left) at ($(A)!-1cm!(B)$);
\coordinate (label_for_A_above) at ($(A)!1cm!($(A)!1cm!90:(O)$)$);
\coordinate (midpoint_on_line_segment_to_position_A) at ($(label_for_A_left)!0.5!(label_for_A_above)$);
\node at ($(A)!3mm!(midpoint_on_line_segment_to_position_A)$){$A$};
\coordinate (label_for_B_right) at ($(B)!-1cm!(A)$);
\coordinate (label_for_B_above) at ($(B)!1cm!($(B)!1cm!-90:(O)$)$);
\coordinate (midpoint_on_line_segment_to_position_B) at ($(label_for_B_right)!0.5!(label_for_B_above)$);
\node at ($(B)!3mm!(midpoint_on_line_segment_to_position_B)$){$B$};
\coordinate (label_for_C_left) at ($(C)!-1cm!(D)$);
\coordinate (label_for_C_below) at ($(C)!1cm!($(C)!1cm!-90:(O)$)$);
\coordinate (midpoint_on_line_segment_to_position_C) at ($(label_for_C_left)!0.5!(label_for_C_below)$);
\node at ($(C)!3mm!(midpoint_on_line_segment_to_position_C)$){$C$};
\coordinate (label_for_D_left) at ($(D)!-1cm!(C)$);
\coordinate (label_for_D_below) at ($(D)!1cm!($(D)!1cm!90:(O)$)$);
\coordinate (midpoint_on_line_segment_to_position_D) at ($(label_for_D_left)!0.5!(label_for_D_below)$);
\node at ($(D)!3mm!(midpoint_on_line_segment_to_position_D)$){$D$};
\draw let \p1=($(B)-(A)$), \n1={atan(\y1/\x1)} in node at (\n1:0.3){$O$};
\draw[dashed] (B) -- (C);
%The angle mark for $\angles{ABC}$ is drawn. It is marked with "|".
\draw[draw=blue] let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(B)!0.5cm!(A)$) arc ({\n1+180}:{\n2+180}:0.5);
\draw[blue] let \p1=($(A)-(B)$), \n1={atan(\y1/\x1)}, \p2=($(B)-(C)$), \n2={atan(\y2/\x2)} in ($(B) +({0.5*(\n1+\n2)+180}:{0.5cm-3pt})$) -- ($(B) +({0.5*(\n1+\n2)+180}:{0.5cm+3pt})$);
\draw[draw=blue] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(D)$), \n2={atan(\y2/\x2)} in ($(C)!0.5cm!(B)$) arc (\n1:\n2:0.5);
\draw[blue] let \p1=($(B)-(C)$), \n1={atan(\y1/\x1)}, \p2=($(C)-(D)$), \n2={atan(\y2/\x2)} in ($(C) +({0.5*(\n1+\n2)}:{0.5cm-3pt})$) -- ($(C) +({0.5*(\n1+\n2)}:{0.5cm+3pt})$);
\end{tikzpicture}
\hspace{\fill}
 A: Some comments on the version as of 2015-10-05-2335 UT:
You have "if" and "only if" reversed.  "If" is $\ell$ is tangent at $P$ if it is perpendicular to the radius $\overline{OP}$.  "Only if" is $\ell$ is perpendicular to the radius $\overline{OP}$ if it is tangent at $P$.  This is just by the way.
Another notational nicety: The distance from $O$ to $P$, which you denote by $|\overline{OP}|$, is more succinctly written just as $OP$.  Context is usually sufficient to determine that this is not the product of $O$ and $P$.
The line containing $P$ that is perpendicular to $\ell$ is variously called $m$ and $k$.
Now on to more substantive matters: Your proof of the "if" direction (labelled "only if" in your proof) is fine, provided that the Pythagorean theorem is available to you.  Your proof of the "only if" direction (tangent $\to$ perpendicular) has the following flaw: You write
$$
OP^2 = OQ^2+PQ^2 < PQ^2
$$
The equality follows from the Pythagorean theorem, all well and good, but how can the inequality be true?  You have a positive number (or at least non-negative), plus another positive number, is less than the second positive number.  Since this is part of your demonstration that if $O$ doesn't lie on $m$, then parts of $\ell$ must be closer to $O$ than $P$ is (violating tangency), this invalidates your proof.
I would argue this direction as follows.  Suppose $\ell$ is not perpendicular to $\overline{OP}$.  Then a perpendicular dropped from $O$ to $\ell$ intersects $\ell$ at some point $R \not= P$.  By the Pythagorean theorem on $\triangle ORP$, we have
$$
OR^2+RP^2 = OP^2
$$
or
$$
OR^2 = OP^2-RP^2 < OP^2
$$
where strict inequality is guaranteed because $R \not= P$, so $RP > 0$.  In other words, there exists a point $R$ on $\ell$ such that $OR < OP$.  But this violates the assumption that $\ell$ is tangent to circle $O$ at $P$.  Therefore, if $\ell$ is tangent to circle $O$ at $P$, it must be perpendicular to $\overline{OP}$.
