construct the center of a given circle ConstructTheCenterOfAGivenCircle 2007-06-08 19:48:27 2007-06-14 17:36:18 Derivation geometric constructions by Euclid % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} \usepackage{pstricks} % there are many more packages, add them here as you need them % define commands here \newcommand{\ol}{\overline} \PMlinkescapeword{center} \PMlinkescapeword{addition} \emph{[Euclid, Book III, Prop. 1]} Find the \PMlinkname{center}{Center8} of a given circle. Since, in Euclidean geometry, a circle has one center only, it suffices to construct a point that is a center of the given circle. Draw any chord $\ol{AB}$ in the circle, and construct the perpendicular bisector of $\ol{AB}$, intersecting $\ol{AB}$ in $C$, and the circle in $D,E$. Let $O$ be the center of the circle; we will show that $O$ is the midpoint of $\ol{DE}$. Note that in the diagram below, $O$ is purposely drawn not to lie on $\ol{DE}$; the proof shows that this position is impossible and that in fact $O$ lies on $\ol{DE}$. It then follows easily that in fact $O$ is the midpoint of $\ol{DE}$. \begin{center} % Generated by eukleides 1.0.3 %--eukleides %frame(-2.8,-2.8,2.6,2.6) %c=circle(point(0,0),2); draw(c) %A=point(c,230:); draw(A) %B=point(c,330:); draw(B) %l=segment(A,B); draw(l) %b=bisector(l); draw(b,dotted) %D E intersection(b,c); draw(D); draw(E); draw(segment(D,E)) %C=intersection(b,line(A,B)); draw(C) %mark(A,C,E,right,0.75) %mark(segment(A,C),double) %mark(segment(C,B),double) %draw("$C$",C,135:); draw("$A$",A,230:); draw("$B$",B,330:); draw("$D$",D,315:); draw("$E$",E,45:) %O=point(0.6,0.5); draw(O); draw("$O$",O,45:) %draw(segment(O,A)); mark(segment(O,A),simple) %draw(segment(O,B)); mark(segment(O,B),simple) %draw(segment(O,C)) %--end \begin{pspicture*}(-2.8000,-2.8000)(2.6000,2.6000) \rput(-2.81,-2.81){.} \rput(2.61,2.61){.} \pscircle(0.0000,0.0000){2.0000} \psdots[dotstyle=*, dotscale=1.0000](-1.2856,-1.5321) \psdots[dotstyle=*, dotscale=1.0000](1.7321,-1.0000) \psline(-1.2856,-1.5321)(1.7321,-1.0000) \psline[linestyle=dotted](0.4585,-2.6000)(-0.4585,2.6000) \psdots[dotstyle=*, dotscale=1.0000](0.3473,-1.9696) \psdots[dotstyle=*, dotscale=1.0000](-0.3473,1.9696) \psline(0.3473,-1.9696)(-0.3473,1.9696) \psdots[dotstyle=*, dotscale=1.0000](0.2232,-1.2660) \psline(0.0017,-1.3051)(-0.0374,-1.0835)(0.1842,-1.0445) \psline(-0.5868,-1.2566)(-0.5347,-1.5520) \psline(-0.5277,-1.2461)(-0.4756,-1.5416) \psline(0.9221,-0.9905)(0.9741,-1.2860) \psline(0.9811,-0.9801)(1.0332,-1.2755) \uput{0.3000}[135.0000](0.2232,-1.2660){$C$} \uput{0.3000}[230.0000](-1.2856,-1.5321){$A$} \uput{0.3000}[330.0000](1.7321,-1.0000){$B$} \uput{0.3000}[315.0000](0.3473,-1.9696){$D$} \uput{0.3000}[45.0000](-0.3473,1.9696){$E$} \psdots[dotstyle=*, dotscale=1.0000](0.6000,0.5000) \uput{0.3000}[45.0000](0.6000,0.5000){$O$} \psline(0.6000,0.5000)(-1.2856,-1.5321) \psline(-0.2328,-0.6181)(-0.4527,-0.4140) \psline(0.6000,0.5000)(1.7321,-1.0000) \psline(1.2858,-0.1596)(1.0463,-0.3404) \psline(0.6000,0.5000)(0.2232,-1.2660) \end{pspicture*} % End of figure \end{center} Since $O$ is the center of the circle, it follows that $OA=OB$. Since $\ol{DE}$ bisects $\ol{AB}$, we see in addition that $AC=BC$. $\triangle ACO$ and $\triangle BCO$ share their third side, $\ol{OC}$. So by SSS, $\triangle ACO \cong \triangle BCO$, and thus, using CPCTC, $\angle ACO\cong\angle BCO$. But $\angle ACO+\angle BCO=180^{\circ}$, so $\angle ACO$ and $\angle BCO$ are each right angles. Thus $O$ in fact lies on $\ol{DE}$. However, since $O$ is the center of the circle, it must be equidistant from $D$ and $E$, and thus $O$ is the midpoint of $\ol{DE}$.