Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

8.1: Circumcircle and circumcenter

  • Page ID
    23626
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Theorem \(\PageIndex{1}\)

    Perpendicular bisectors to the sides of any nondegener- ate triangle intersect at one point.

    The point of intersection of the perpendicular bisectors is called circumcenter. It is the center of the circumcircle of the triangle; that is, a circle that passes thru all three vertices of the triangle. The circumcenter of the triangle is usually denoted by \(O\).

    Proof

    Let \(\triangle ABC\) be nondegenerate. Let \(\ell\) and \(m\) be perpendicular bisectors to sides \([AB]\) and \([AC]\) respectively.

    Assume \(\ell\) and \(m\) intersect, let \(O = \ell \cap m\).

    Let us apply Theorem 5.2.1. Since \(O \in \ell\), we have that \(OA = OB\) and since \(O \in m\), we have that \(OA = OC\). It follows that \(OB = OC\); that is, \(O\) lies on the perpendicular bisector to \([BC]\).

    It remains to show that \(\ell \nparallel m\); assume the contrary. Since \(\ell \perp (AB)\) and \(m \perp (AC)\), we get that \((AC) \parallel (AB)\) (see Exercise 7.1.1). Therefore, by Theorem 5.3.1, \((AC) = (AB)\); that is, \(\triangle ABC\) is degenerate - a contradiction.

    Exercise \(\PageIndex{1}\)

    There is a unique circle that passes thru the vertexes of a given nondegenerate triangle in the Euclidean plane.

    Answer

    Apply Theorem \(\PageIndex{1}\) and Theorem 5.2.1.