How to draw a tangent line for a circle which come through a given point outside of the circle

Question

this is my code,

\begin{tikzpicture}

\node (A) at (-4, -3) {A};

\node (o) at (0, 0) {o};

\draw [name path = circle] (o) circle (3.5cm); 

\end{tikzpicture}

Given a point A outside a circle, I want to:

draw a tangent line to the circle which come through point A.
find the intersection point (between the tangent line and the circle) and denoted this point as B. (I need to use this point to connect to another point later).

Is there a way to do it? Thanks. :)

see tikz & pgf manual, example on the top of the page 137. — Zarko, Apr 12 '18 at 22:20
I know that code, it does not give you the intersection point between tangent line and the circle. — Yikun Bai, Apr 12 '18 at 22:30

score 5 · Answer 1 · 2018-04-12T22:43:53.590

UPDATE: OK, sorry for misreading your question. I guess I then agree with @Zarko, this is almost identical to the example from the pgfmanual.

\documentclass[tikz,border=3.14pt]{standalone}
\usetikzlibrary{intersections,calc}
\begin{document}
\begin{tikzpicture}

\node (A) at (-4, -3) {A};

\node (o) at (0, 0) {o};

\node [circle,draw,name path=circle] (c) at (o) [minimum size=7cm] {};


\draw[red] (A)  -- (tangent cs:node=c,point={(A)},solution=1) 
coordinate[label=below:B] (B)
       (A) -- (tangent cs:node=c,point={(A)},solution=2)
       coordinate[label=left:C] (C);
\end{tikzpicture}
\end{document}

\documentclass[tikz,border=3.14pt]{standalone}
\usetikzlibrary{intersections,calc}
\begin{document}
\begin{tikzpicture}

\node (A) at (-4, -3) {A};

\node (o) at (0, 0) {o};

\draw[name path=circle] (o) circle (3.5cm); 
\draw[name path=Ao] (A) to[bend left=0](o);
\path [name intersections={of=Ao and circle,total=\t}]
coordinate[label=below:B] (B) at (intersection-1);
\draw ($ (B)!3cm!90:(A) $) -- ($ (B)!3cm!270:(A) $);
\end{tikzpicture}
\end{document}

Thank you, it should not looks like this, it means the tangent line come through A, not perpendicular to OA. (I do not know how to add a picture yet, the correct result should looks like this, by connecting OBA, it would form a right triangular. :) — Yikun Bai, Apr 12 '18 at 22:34
In the first diagram, OBA does form a right triangle. The radius is always perpendicular to a tangent. — John Kormylo, Apr 13 '18 at 01:53
Hi John Kormylo! I agree. Probably it will become clearer if you address your comment to @Labibi — , Apr 13 '18 at 02:04

score 2 · Answer 2 · answered Apr 12 '18 at 23:48

Here is a way to do the same construction with pst-eucl:

\documentclass[11pt, svgnames, border = 5pt]{standalone}
\usepackage{pstricks-add} % loads also pst-node
\usepackage{pst-eucl} % for plane geometry

% \usepackage{auto-pst-pdf} % to compile with pdflatex --enable-write18 (MiKTeX) or pdflatex --shell-escape (TX Live, MacTeX))

\begin{document}

\begin{pspicture}(-6,-5)(4,4)
\psset{PointSymbol=none, dotsize=2.5pt, linejoin=1, dimen=outer, unit=1cm}
\pstGeonode[PosAngle={30,-150}](0, 0){O}(-4,-3){A}
\pstCircleOA[Radius =\pstDistVal{3.5}, linecolor = IndianRed, linewidth = 1.2pt]{O}{}
\pstMiddleAB[ PointName=none]{O}{A}{I}
 \psset{linewidth=0.6pt}
 \pstInterCC[RadiusA=\pstDistVal{3.5}, DiameterB=\pstDistAB{O}{A},
 CodeFigB=true, CodeFigColor=Gold, PointName=default]{O}{}{I}{}{C}{B}
 \uput[l](B){B} \uput[d](C){C}
 \psset{linecolor=Tomato, nodesep=-2}
 \pstLineAB{A}{B}\pstLineAB{A}{C}
 \psline[linestyle=dashed](B)(O)(C)
 \psset{linecolor = LightSteelBlue, linewidth=0.4pt, RightAngleSize=0.15}
 \pstRightAngle*{A}{B}{O}
 \pstRightAngle{A}{C}{O}
\end{pspicture}

\end{document}

How to draw a tangent line for a circle which come through a given point outside of the circle

2 Answers2