How to draw a line parallel to diameter intersecting the circumference of a circle

Question

I have drawn a circle and its horizontal and vertical line.

\begin{tikzpicture}

  \draw (0,0) node [above left] {O} circle (3); % circle of radius 3
  \draw (0,3) node [above] {A} -- (0,-3) node [below] {B}; % vertical diameter
  \draw (-3,0) node [left] {P} -- (3,0) node [right] {Q}; % hotizontal diameter

\end{tikzpicture}

Now, I want a line parallel to AB passing through (-2,0) and ending on the circumference of the circle at both ends.

Answer 1

Two different approaches:

Clipping

You can draw a line longer than required, but clipping it to the circle:

\begin{tikzpicture}
  \draw (0,0) node [above left] {O} circle (3); % circle of radius 3
  \draw (0,3) node [above] {A} -- (0,-3) node [below] {B}; % vertical diameter
  \draw (-3,0) node [left] {P} -- (3,0) node [right] {Q}; % hotizontal diameter
\clip (0,0) circle (3);
\draw[green] (-2,3) -- (-2,-3);
\end{tikzpicture}

Calc

You can use trigonometry to calculate the coordinates of the starting and ending points of the line. Using polar coordinates, what you have to find are the angles of those points with respect of the center of the circle. In this case it is easy, since the radius r and the x distance are known, the angle will be acos(x/r). This calculation can be made in a let..in construct.

With this approach you have also the coordinates (in polar form) of the end points of the line, so you can label them also if you need to (as shown in the following example).

\usetikzlibrary{calc}
\begin{tikzpicture}
  \draw (0,0) node [above left] {O} circle (3); % circle of radius 3
  \draw (0,3) node [above] {A} -- (0,-3) node [below] {B}; % vertical diameter
  \draw (-3,0) node [left] {P} -- (3,0) node [right] {Q}; % hotizontal diameter

\draw[red] 
   let \n1 = {acos(2/3)}        % \n1 is the angle
   in (180-\n1:3) node[above left] {A'} -- 
      (180+\n1:3) node[below left] {B'};
   % Polar form uses the syntax (angle:radius)
\end{tikzpicture}

Answer 2

You can avoid the math with the intersections library of tikz:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{intersections}
\begin{document}

\begin{tikzpicture}
  \draw[name path=circle] (0,0) circle (3) node [above left] {$O$};  % circle of radius 3
  \draw (0,3) node [above] {$A$} -- (0,-3) node [below] {$B$}; % vertical diameter
  \draw (-3,0) node [left] {$P$} -- (3,0) node [right] {$Q$}; % horizontal diameter
  \path[name path=transversal] (-2,-3) -- ++(0,6);
  \draw[name intersections={of=circle and transversal,by={Q',P'}}] 
     (P') node[below] {$P'$} -- (Q') node[above] {$Q'$};
\end{tikzpicture}

\end{document}

This is a pretty flexible approach because the paths can be really anything—circles, lines, Bézier curves, etc.

Answer 3

With a little computation, you can get the angle and then get the symmetrical point.

\begin{tikzpicture}

  \draw (0,0) node [above left] {O} circle (3); % circle of radius 3
  \draw (0,3) node [above] {A} -- (0,-3) node [below] {B}; % vertical diameter
  \draw (-3,0) node [left] {P} -- (3,0) node [right] {Q}; % hotizontal diameter

\draw ({asin(-2/3)}:3) -- ({-180-asin(-2/3)}:3);
\end{tikzpicture}

Answer 4

Good job for tkz-euclide. You have a lot of solutions with tikz, the choice depends of what you want to do after this. The solution with tkz-euclide is interesting if you want to avoid some calculations, if you want to draw a lot of pictures like this. The only problem comes from my macro \tkzInterLC because in some cases the result is not accurate enough. Interesting too : you can easily adapt the solution for different radius.

\documentclass{article}
\usepackage{tkz-euclide}
\usetkzobj{all}

\begin{document}
\begin{tikzpicture}
 \tkzInit[ymin=-4,ymax=4,xmin=-4,xmax=4]
% definitions
 \tkzDefPoints{ 0  / 0 /O, %blank spaces are fine for numbers 
                3  / 0 /P, % but not for the names of nodes
                3  / 0 /Q,
                0  / 3 /A,
                0  /-3 /B,
               -2  / 0 /I}
\tkzDefPointWith[orthogonal](I,O) \tkzGetPoint{H} % IH perp IO
\tkzInterLC(I,H)(O,A)         \tkzGetPoints{E}{F} % inter circle and line
% drawings
\tkzDrawCircle[R](O,3 cm)
\tkzDrawSegments(A,B P,Q)
\tkzDrawSegment[color=red](E,F)
\tkzDrawPoints(E,F)
% labels
\tkzLabelPoints(O,P,Q,A,B,I)
\end{tikzpicture}
\end{document}  

If you want only a line parallel and through a point, possible is

\tkzDefLine[parallel=through I](A,B) \tkzGetPoint{J}

Answer 5

With PSTricks.

Intersection

\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pst-eucl}

\psset{CurveType=polyline}

\begin{document}
\begin{pspicture}(-3,-3)(3,3)
    \pstGeonode[PosAngle={135,0,180}]{O}(3,0){Q}(-3,0){P}
    \pstGeonode[PosAngle={90,-90}](0,3){A}(0,-3){B}
    \pnode(-2,0|A){A'}
    \pnode(-2,0|B){B'}
    \pstInterLC[PosAngleA=90,PosAngleB=-90]{A'}{B'}{O}{A}{Q'}{P'}
    \pstCircleOA{O}{A}
    \psline(Q')(P')
\end{pspicture}
\end{document}

Warning: I just knew that

    \pstInterLC[PosAngle={90,-90}]{A'}{B'}{O}{A}{Q'}{P'}

will not work!

Inverse Trigonometry

As PostScript only provides sin, cos, and atan, we need to define a new RPN operator acos as

\pstVerb{/acos {dup 2 exp 1 exch sub sqrt exch atan} bind def}

The complete code is as follows.

\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pst-eucl}

\pstVerb{/acos {dup 2 exp 1 exch sub sqrt exch atan} bind def}

\psset{CurveType=polyline}

\begin{document}
\begin{pspicture}(-3,-3)(3,3)
    \pstGeonode[PosAngle={135,0,180}]{O}(3,0){Q}(-3,0){P}
    \pstGeonode[PosAngle={90,-90}](0,3){A}(0,-3){B}
    \pstGeonode[PosAngle={90,-90}]
        (!3 -2 3 div acos PtoC){Q'}
        (!3 -2 3 div acos neg PtoC){P'}
    \pstCircleOA{O}{A}
\end{pspicture}
\end{document}

Clipping

Using clipping (for this problem) should be avoided because it makes us difficult to place the P' and Q' labels (and the dots if needed).

The following example uses node P' and Q' (used in the second method above) to circumvent the difficulty but it will look so funny because of

    \pstGeonode[PosAngle={90,-90},CurveType=none]
        (!3 -2 3 div acos PtoC){Q'}
        (!3 -2 3 div acos neg PtoC){P'
    \psclip{\pstCircleOA{O}{A}}
        \psline([offset=1]Q')([offset=-1]P')
    \endpsclip

which can actually be written as

    \pstGeonode[PosAngle={90,-90}]
        (!3 -2 3 div acos PtoC){Q'}
        (!3 -2 3 div acos neg PtoC){P'}

OK, here is the complete code!

\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pst-eucl}

\pstVerb{/acos {dup 2 exp 1 exch sub sqrt exch atan} bind def}

\psset{CurveType=polyline}

\begin{document}
\begin{pspicture}(-3,-3)(3,3)
    \pstGeonode[PosAngle={135,0,180}]{O}(3,0){Q}(-3,0){P}
    \pstGeonode[PosAngle={90,-90}](0,3){A}(0,-3){B}
    \pstGeonode[PosAngle={90,-90},CurveType=none]
        (!3 -2 3 div acos PtoC){Q'}
        (!3 -2 3 div acos neg PtoC){P'}
    \psclip{\pstCircleOA{O}{A}}
        \psline([offset=1]Q')([offset=-1]P')
    \endpsclip
    %\pstCircleOA{O}{A} <-- not needed!
\end{pspicture}
\end{document}

The latest edit:

Apparently, pstricks.pro has added or defined Acos as follows,

/Acos {dup dup mul neg 1 add dup 0 lt {% arc cos, returns 0 when negative root
  pop pop 0 }{ sqrt exch atan} ifelse } def

so more keystrokes can be reduced.

\documentclass[pstricks,border=15pt]{standalone}
\usepackage{pst-eucl}
\psset{CurveType=polyline}

\begin{document}
\begin{pspicture}(-3,-3)(3,3)
    \pstGeonode[PosAngle={135,0,180}]{O}(3,0){Q}(-3,0){P}
    \pstGeonode[PosAngle={90,-90}](0,3){A}(0,-3){B}
    \pstGeonode[PosAngle={90,-90}]
        (!3 -2 3 div Acos PtoC){Q'}
        (!3 -2 3 div Acos neg PtoC){P'}
    \pstCircleOA{O}{A}
\end{pspicture}
\end{document}