\section{Intersections} You can determine the points of intersection of two straight lines, of a straight line and a circle, and of two circles. The associated commands do not check that the intersection points exist: this is up to the user (see the tests \tkzcname{tkzTestInterLC} and \tkzcname{tkzTestInterCC}). \subsection{Intersection of two straight lines \tkzcname{tkzInterLL}} \begin{NewMacroBox}{tkzInterLL}{\parg{$A,B$}\parg{$C,D$}}% Defines the intersection point \tkzname{tkzPointResult} of the two lines $(AB)$ and $(CD)$. The known points are given in pairs (two per line) between parentheses, and the resulting point is retrieved with the macro \tkzcname{tkzGetPoint}. \end{NewMacroBox} \subsubsection{Example of intersection between two straight lines} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[rotate=-45,scale=.75] \tkzDefPoint(2,1){A} \tkzDefPoint(6,5){B} \tkzDefPoint(3,6){C} \tkzDefPoint(5,2){D} \tkzDrawLines(A,B C,D) \tkzInterLL(A,B)(C,D) \tkzGetPoint{I} \tkzDrawPoints[color=blue](A,B,C,D) \tkzDrawPoint[color=red](I) \end{tikzpicture} \end{tkzexample} \subsection{Intersection of a straight line and a circle \tkzcname{tkzInterLC}} As before, the line is defined by a pair of points. The circle is also defined by a pair: \begin{itemize} \item $(O,C)$, a pair of points: the first one is the center and the second one is any point of the circle; \item $(O,r)$, where $r$ is the radius. \end{itemize} \begin{NewMacroBox}{tkzInterLC}{\oarg{options}\parg{$A,B$}\parg{$O,C$} or \parg{$O,r$} or \parg{$O,C,D$}}% So the arguments are two pairs. \medskip \begin{tabular}{lll}% \toprule options & default & definition \\ \midrule \TOline{N} {N} {(O,C) determines the circle} \TOline{R} {N} {(O, 1 ) unit 1 cm} \TOline{with nodes}{N} {(O,C,D) CD is a radius} \TOline{common=pt} {} {pt is common point; tkzFirstPoint gives the other point} \TOline{near} {} {tkzFirstPoint is the closest point to the first point of the line} \bottomrule \end{tabular} \medskip The macro defines the intersection points $I$ and $J$ of the line $(AB)$ and the circle with center $O$ and radius $r$, if they exist; otherwise, an error is reported in the |.log| file. With \tkzname{with nodes}, you don't need to calculate the radius, which is the length of $[CD]$. If \tkzname{common} and \tkzname{near} are not used, \tkzname{tkzFirstPoint} is the point that gives the smallest angle (formed with \tkzname{tkzSecondPoint} and the center of the circle). \end{NewMacroBox} \begin{NewMacroBox}{tkzTestInterLC}{\parg{$O,A$}\parg{$O',B$}}% The arguments are two pairs, which define a line and a circle (with a center and a point of the circle). If the line and the circle intersect, the test \tkzcname{iftkzFlagLC} is true. \end{NewMacroBox} \subsubsection{Testing a line-circle intersection: \tkzcname{tkzTestInterLC}} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=1] \tkzDefPoints{% x y name 3 /4 /I, 3 /2 /P, 0 /2 /La, 8 /3 /Lb} \tkzDrawCircle(I,P) \foreach \i in {1,...,3}{% \coordinate (Lb) at (8,\i); \tkzDrawLine(La,Lb) \tkzTestInterLC(La,Lb)(I,P) \iftkzFlagLC \tkzInterLC(La,Lb)(I,P) \tkzGetPoints{a}{b} \tkzDrawPoints(a,b) \fi } \end{tikzpicture} \end{tkzexample} \subsubsection{Line-circle intersection} In the following example, the circle is drawn with two points, and the intersection of the straight line and the circle uses two pairs of points. We will compare the angles $\widehat{D,E,O}$ and $\widehat{E,D,O}$. These angles have opposite orientations. \tkzname{tkzFirstPoint} is assigned to the point that forms the angle with the smallest measure (counterclockwise direction). The counterclockwise angle $\widehat{D,E,O}$ measures $360^\circ$ minus the measure of $\widehat{O,E,D}$. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.75] \tkzInit[xmax=5,ymax=4] \tkzDefPoint(1,1){O} \tkzDefPoint(-2,4){La} \tkzDefPoint(5,0){Lb} \tkzDefPoint(3,3){C} \tkzInterLC(La,Lb)(O,C) \tkzGetPoints{D}{E} \tkzMarkAngle[->,size=1.5](E,D,O) \tkzDrawPolygons[new](O,D,E) \tkzMarkAngle[->,size=1.5](D,E,O) \tkzDrawCircle(O,C) \tkzDrawPoints[color=teal](O,La,Lb,C) \tkzDrawPoints[color=red](D,E) \tkzDrawLine(La,Lb) \tkzLabelPoints[above right](O,La,Lb,C,D,E) \end{tikzpicture} \end{tkzexample} \subsubsection{Line passing through the center: option \tkzname{common}} This case is special: the angles cannot be compared, so the option \tkzname{near} must be used. \tkzname{tkzFirstPoint} is assigned to the point closest to the first point given for the line. Here we want $A$ to be closest to $Lb$. \begin{tkzexample}[latex=8cm,small] \begin{tikzpicture} \tkzDefPoints{% x y name 0 /1 /D, 6 /0 /B, 3 /3 /O, 2 /2 /La, 5 /5 /Lb} \tkzDrawCircle(O,D) \tkzDrawLine(La,Lb) \tkzInterLC[near](Lb,La)(O,D) \tkzGetFirstPoint{A} \tkzDrawSegments(O,A) \tkzDrawPoints(O,D,A,La,Lb) \tkzLabelPoints(O,D,A,La,Lb) \end{tikzpicture} \end{tkzexample} \subsubsection{Line-circle intersection with option \tkzname{common}} A special case that often occurs: a point of the line is on the circle, and we look for the other common point. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoints{0/0/O,-5/0/A,2/-2/B,0/5/D} \tkzInterLC[common=A](B,A)(O,D) \tkzGetFirstPoint{C} \tkzDrawPoints(O,A,B) \tkzDrawCircle(O,A) \tkzDrawLine(A,C) \tkzDrawPoint(C) \tkzLabelPoints(A,B,C) \end{tikzpicture} \end{tkzexample} \subsubsection{Line-circle intersection: order of the points} The idea is to compare the angles formed by the first point of the line, a resulting point and the center of the circle. The first point is the one that gives the smallest angle. As you can see, $\widehat{BCO} < \widehat{BEO}$; in fact, $\widehat{BEO}$ is counterclockwise. \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoints{0/0/O,5/1/A,2/2/B,3/1/D} \tkzInterLC[common=A](B,D)(O,A) \tkzGetPoints{C}{E} \tkzDrawPoints(O,A,B,D) \tkzDrawCircle(O,A) \tkzDrawLine(E,C) \tkzDrawSegments[dashed](B,O O,C) \tkzMarkAngle[->,size=1.5](B,C,O) \tkzDrawSegments[dashed](O,E) \tkzMarkAngle[->,size=1.5](B,E,O) \tkzDrawPoints(C,E) \tkzLabelPoints[above](O,E) \tkzLabelPoints[right](A,B,C,D) \end{tikzpicture} \end{tkzexample} \subsubsection{Example with \tkzcname{foreach}} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=3,rotate=180] \tkzDefPoint(0,1){J} \tkzDefPoint(0,0){O} \foreach \i in {0,-5,-10,...,-90}{ \tkzDefPoint({2.5*cos(\i*pi/180)},% {1+2.5*sin(\i*pi/180)}){P} \tkzInterLC[R](P,J)(O,1)\tkzGetPoints{N}{M} \tkzDrawSegment[color=orange](J,N) \tkzDrawPoints[red](N)} \foreach \i in {-90,-95,...,-175,-180}{ \tkzDefPoint({2.5*cos(\i*pi/180)},% {1+2.5*sin(\i*pi/180)}){P} \tkzInterLC[R](P,J)(O,1)\tkzGetPoints{N}{M} \tkzDrawSegment[color=orange](J,M) \tkzDrawPoints[red](M)} \end{tikzpicture} \end{tkzexample} \subsubsection{Line-circle intersection with option \tkzname{near}} $D$ is the point closest to $b$. \begin{tkzexample}[vbox,small] \begin{tikzpicture} \tkzDefPoints{0/0/A,12/0/C} \tkzDefGoldenRatio(A,C) \tkzGetPoint{B} \tkzDefMidPoint(A,C) \tkzGetPoint{O} \tkzDefMidPoint(A,B) \tkzGetPoint{O_1} \tkzDefMidPoint(B,C) \tkzGetPoint{O_2} \tkzDefPointBy[rotation=center O_2 angle 90](C) \tkzGetPoint{P} \tkzDefPointBy[rotation=center O_1 angle 90](B) \tkzGetPoint{Q} \tkzDefPointBy[rotation=center B angle 90](C) \tkzGetPoint{b} \tkzInterLC[near](b,B)(O,A) \tkzGetFirstPoint{D} \tkzInterCC(D,B)(O,C) \tkzGetPoints{V}{U} \tkzDefPointBy[projection=onto U--V](O_1) \tkzGetPoint{M} \tkzDefPointBy[projection=onto U--V](O_2) \tkzGetPoint{N} \tkzDrawPoints(A,B,C,O,O_1,O_2,D,U,V,M,N,b) \tkzDrawSemiCircles[teal](O,C O_1,B O_2,C) \tkzDrawSegments(A,C B,D U,V A,D C,D M,B B,N) \tkzDrawArc(D,U)(V) \tkzLabelPoints(A,B,C,O,O_1,O_2) \tkzLabelPoints[above](D,U,V,M,N) \end{tikzpicture} \end{tkzexample} \subsubsection{More complex example of a line-circle intersection} Figure from \url{http://gogeometry.com/problem/p190_tangent_circle} \begin{tkzexample}[latex=6.5cm,small] \begin{tikzpicture}[scale=.75] \tkzDefPoint(0,0){A} \tkzDefPoint(8,0){B} \tkzDefMidPoint(A,B) \tkzGetPoint{O} \tkzDefMidPoint(O,B) \tkzGetPoint{O'} \tkzDefLine[tangent from=A](O',B)\tkzGetFirstPoint{E} \tkzInterLC(A,E)(O,B) \tkzGetFirstPoint{D} \tkzDefPointBy[projection=onto A--B](D) \tkzGetPoint{F} \tkzDrawCircles(O,B O',B) \tkzDrawSegments(A,D A,B D,F) \tkzDrawSegments[color=red,line width=1pt, opacity=.4](A,O F,B) \tkzDrawPoints(A,B,O,O',E,D) \tkzMarkRightAngle(D,F,B) \tkzLabelPoints[below right](A,B,O,O',E,D) \end{tikzpicture} \end{tkzexample} \subsubsection{Circle defined by a center and a measure, and special cases} Let's look at some special cases, such as straight lines tangent to the circle. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoint(0,8){A} \tkzDefPoint(8,0){B} \tkzDefPoint(8,8){C} \tkzDefPoint(4,4){D} \tkzDefPoint(2,4){E} \tkzDefPoint(4,2){F} \tkzDefPoint(8,4){G} \tkzInterLC(A,C)(D,G) \tkzGetPoints{I1}{I2} \tkzInterLC(B,C)(D,G) \tkzGetPoints{J1}{J2} \tkzInterLC[near](A,B)(D,G) \tkzGetPoints{K1}{K2} \tkzInterLC(E,F)(D,G) \tkzGetPoints{E1}{E2} \tkzDrawCircle(D,G) \tkzDrawPoints[color=red](I1,J1,K1,K2,E1,E2) \tkzDrawLines(A,B B,C A,C I2,J2 E1,E2) \tkzDrawPoints[color=blue](A,...,F) \tkzDrawPoints[color=red](I2,J2) \tkzLabelPoints[left](B,D,E,F) \tkzLabelPoints[below left](A,C) \tkzLabelPoints[below=4pt](I1,K1,K2,E2) \tkzLabelPoints[left](J1,E1) \end{tikzpicture} \end{tkzexample} \subsubsection{Calculation of radius} With \tkzname{pgfmath} and \tkzcname{pgfmathsetmacro} The radius may be the result of a calculation, done before the intersection macro rather than inside it. A length can be calculated in several ways. You can of course use the module \tkzname{pgfmath} and the macro \tkzcname{pgfmathsetmacro}. In some cases the results are not precise enough: the calculation $0.0002 \div 0.0001$ gives $1.98$ with pgfmath, while xfp gives $2$. With \tkzname{xfp} and \tkzcname{fpeval}: \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture} \tkzDefPoint(2,2){A} \tkzDefPoint(5,4){B} \tkzDefPoint(4,4){O} \pgfmathsetmacro\tkzLen{\fpeval{0.0002/0.0001}} % or \edef\tkzLen{\fpeval{0.0002/0.0001}} \tkzInterLC[R](A,B)(O, \tkzLen) \tkzGetPoints{I}{J} \tkzDrawCircle(O,I) \tkzDrawPoints[color=blue](A,B) \tkzDrawPoints[color=red](I,J) \tkzDrawLine(I,J) \end{tikzpicture} \end{tkzexample} \subsubsection{Option \code{with nodes}} \begin{tkzexample}[latex=8cm,small] \begin{tikzpicture}[scale=.75] \tkzDefPoints{0/0/A,4/0/B,1/1/D,2/0/E} \tkzDefTriangle[equilateral](A,B) \tkzGetPoint{C} \tkzInterLC[with nodes](D,E)(C,A,B) \tkzGetPoints{F}{G} \tkzDrawCircle(C,A) \tkzDrawPolygon(A,B,C) \tkzDrawPoints(A,...,G) \tkzDrawLine(F,G) \end{tikzpicture} \end{tkzexample} \subsection{Intersection of two circles \tkzcname{tkzInterCC}} The most frequent case is that of two circles defined by their center and a point, but as before the option \tkzname{R} allows you to give the radius as a measure. \begin{NewMacroBox}{tkzInterCC}{\oarg{options}\parg{$O,A$}\parg{$O',A'$} or \parg{$O,r$}\parg{$O',r'$} or \parg{$O,A,B$} \parg{$O',C,D$}}% \begin{tabular}{lll}% options & default & definition \\ \midrule \TOline{N} {N} {$OA$ and $O'A'$ are radii, $O$ and $O'$ are the centers.} \TOline{R} {N} {$r$ and $r'$ are dimensions and measure the radii.} \TOline{with nodes} {N} {in (A,A,C)(C,B,F) AC and BF give the radii. } \TOline{common=pt} {} {pt is common point; tkzFirstPoint gives the other point.} \bottomrule \end{tabular} \medskip This macro defines the intersection points $I$ and $J$ of the two circles with centers $O$ and $O'$. If the two circles have no common point, the macro ends with an error that is not handled. If the centers are $O$ and $O'$ and the intersections are $A$ and $B$, the angles $\widehat{O,A,O'}$ and $\widehat{O,B,O'}$ have opposite orientations; \tkzname{tkzFirstPoint} is assigned to the point that forms the \code{clockwise} angle. \end{NewMacroBox} \begin{NewMacroBox}{tkzTestInterCC}{\parg{$O,A$}\parg{$O',B$}}% The arguments are two pairs, which define two circles (each with a center and a point of the circle). If the two circles intersect, the test \tkzcname{iftkzFlagCC} is true. \end{NewMacroBox} \subsubsection{Testing a circle-circle intersection: \tkzcname{tkzTestInterCC}} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.75] \tkzDefPoints{% x y name 0 /0 /A, 2 /0 /B, 4 /0 /I, 1 /0 /P} \tkzDrawCircle(A,B) \foreach \i in {1,...,3}{% \coordinate (P) at (\i,0); \tkzDrawCircle[new](I,P) \tkzTestInterCC(A,B)(I,P) \iftkzFlagCC \tkzInterCC(A,B)(I,P) \tkzGetPoints{a}{b} \tkzDrawPoints(a,b) \fi} \end{tikzpicture} \end{tkzexample} \subsubsection{Circle-circle intersection with a \tkzname{common} point} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoints{0/0/O,5/-1/A,2/2/B} \tkzDrawPoints(O,A,B) \tkzDrawCircles(O,B A,B) \tkzInterCC[common=B](O,B)(A,B) \tkzGetFirstPoint{C} \tkzDrawPoint(C) \tkzLabelPoints[above](O,A,B,C) \end{tikzpicture} \end{tkzexample} \subsubsection{Circle-circle intersection: order of the points} The idea is to compare the angles formed by the first center, a resulting point and the center of the second circle. The first point is the one that gives the smallest angle. As you can see, $\widehat{ODB} < \widehat{OBE}$. \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture}[scale=.5] \pgfkeys{/pgf/number format/.cd,fixed relative, precision=4} \tkzDefPoints{0/0/O,5/-1/A,2/2/B,2/-1/C} \tkzDrawPoints(O,A,B) \tkzDrawCircles(O,A B,C) \tkzInterCC(O,A)(B,C)\tkzGetPoints{D}{E} \tkzDrawPoints(C,D,E) \tkzLabelPoints(O,A,B,C) \tkzLabelPoints[above](D,E) \tkzDrawSegments[cyan](D,O D,B) \tkzMarkAngle[red,->,size=1.5](O,D,B) \tkzFindAngle(O,D,B) \tkzGetAngle{an} \tkzLabelAngle(O,D,B){$ \pgfmathprintnumber{\an}$} \tkzDrawSegments[cyan](E,O E,B) \tkzMarkAngle[red,->,size=1.5](O,E,B) \tkzFindAngle(O,E,B) \tkzGetAngle{an} \tkzLabelAngle(O,E,B){$ \pgfmathprintnumber{\an}$} \end{tikzpicture} \end{tkzexample} \subsubsection{Construction of an equilateral triangle} $\widehat{A,C,B}$ is a clockwise angle. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[trim left=-1cm,scale=.5] \tkzDefPoint(1,1){A} \tkzDefPoint(5,1){B} \tkzInterCC(A,B)(B,A)\tkzGetPoints{C}{D} \tkzDrawPoint[color=black](C) \tkzDrawCircles(A,B B,A) \tkzCompass[color=red](A,C) \tkzCompass[color=red](B,C) \tkzDrawPolygon(A,B,C) \tkzMarkSegments[mark=s|](A,C B,C) \tkzLabelPoints[](A,B) \tkzLabelPoint[above](C){$C$} \end{tikzpicture} \end{tkzexample} \subsubsection{Segment trisection} The idea here is to divide a segment into three equal parts with a ruler and a compass. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.6] \tkzDefPoint(0,0){A} \tkzDefPoint(3,2){B} \tkzInterCC(A,B)(B,A) \tkzGetSecondPoint{D} \tkzInterCC(D,B)(B,A) \tkzGetPoints{A}{C} \tkzInterCC(D,B)(A,B) \tkzGetPoints{E}{B} \tkzInterLC[common=D](C,D)(E,D)\tkzGetFirstPoint{F} \tkzInterLL(A,F)(B,C) \tkzGetPoint{O} \tkzInterLL(O,D)(A,B) \tkzGetPoint{H} \tkzInterLL(O,E)(A,B) \tkzGetPoint{G} \tkzDrawCircles(D,E A,B B,A E,A) \tkzDrawSegments[](O,F O,B O,D O,E) \tkzDrawPoints(A,...,H) \tkzDrawSegments(A,B B,D A,D A,E E,F C,F B,C) \tkzMarkSegments[mark=s|](A,G G,H H,B) \end{tikzpicture} \end{tkzexample} \subsubsection{With the option \code{with nodes}} \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoints{0/0/A,0/5/B,5/0/C} \tkzDefPoint(54:5){F} \tkzInterCC[with nodes](A,A,C)(C,B,F) \tkzGetPoints{a}{e} \tkzInterCC(A,C)(a,e) \tkzGetFirstPoint{b} \tkzInterCC(A,C)(b,a) \tkzGetFirstPoint{c} \tkzInterCC(A,C)(c,b) \tkzGetFirstPoint{d} \tkzDrawCircle[new](A,C) \tkzDrawPoints(a,b,c,d,e) \tkzDrawPolygon(a,b,c,d,e) \foreach \vertex/\num in {a/36,b/108,c/180, d/252,e/324}{% \tkzDrawPoint(\vertex) \tkzLabelPoint[label=\num:$\vertex$](\vertex){} \tkzDrawSegment(A,\vertex) } \end{tikzpicture} \end{tkzexample} \subsubsection{Mix of intersections} \begin{tkzexample}[latex=8cm,small] \begin{tikzpicture}[scale = .7] \tkzDefPoint(2,2){A} \tkzDefPoint(0,0){B} \tkzDefPoint(-2,2){C} \tkzDefPoint(0,4){D} \tkzDefPoint(4,2){E} \tkzCircumCenter(A,B,C)\tkzGetPoint{O} \tkzInterCC[R](O,2)(D,2)\tkzGetPoints{M1}{M2} \tkzInterCC(O,A)(D,O) \tkzGetPoints{1}{2} \tkzInterLC(A,E)(B,M1)\tkzGetSecondPoint{M3} \tkzInterLC(O,C)(M3,D)\tkzGetSecondPoint{L} \tkzDrawSegments(C,L) \tkzDrawPoints(A,B,C,D,E,M1,M2,M3,O,L) \tkzDrawSegments(O,E) \tkzDrawSegments[new](C,A D,B) \tkzDrawPoint(O) \tkzDrawCircles[new](M3,D B,M2 D,O) \tkzDrawCircle(O,A) \tkzLabelPoints[below right](A,B,C,D,E,M1,M2,M3,O,L) \end{tikzpicture} \end{tkzexample} \subsubsection{Altshiller-Court's theorem} The two lines joining the points of intersection of two orthogonal circles to a point of one of the circles meet the other circle in two diametrically opposite points (Altshiller-Court, p.~176). \begin{tkzexample}[vbox,small] \begin{tikzpicture} \tkzDefPoints{0/0/P,5/0/Q,3/2/I} \tkzDefCircle[orthogonal from=P](Q,I) \tkzGetFirstPoint{E} \tkzDrawCircles(P,E Q,E) \tkzInterCC[common=E](P,E)(Q,E) \tkzGetFirstPoint{F} \tkzDefPointOnCircle[through = center P angle 80 point E] \tkzGetPoint{A} \tkzInterLC[common=E](A,E)(Q,E) \tkzGetFirstPoint{C} \tkzInterLL(A,F)(C,Q) \tkzGetPoint{D} \tkzDrawLines[add=0 and 1](P,Q) \tkzDrawLines[add=0 and 2](A,E) \tkzDrawSegments(P,E E,F F,C A,F C,D) \tkzDrawPoints(P,Q,E,F,A,C,D) \tkzLabelPoints(P,Q,F) \tkzLabelPoints[above](E,A) \tkzLabelPoints[left](D) \tkzLabelPoints[above right](C) \end{tikzpicture} \end{tkzexample} \endinput