Dreieck/Aus normeuklidisch/Umkreismittelpunkt/Aufgabe/Lösung

${\displaystyle {}x={\frac {1}{2}}}$

${\displaystyle {}{\left({\frac {1}{4}},{\frac {\sqrt {\vert {D}\vert }}{4}}\right)}+t{\left({\sqrt {\vert {D}\vert }},-1\right)}}$

${\displaystyle {\frac {1}{4}}+t{\sqrt {\vert {D}\vert }}={\frac {1}{2}}\,\,{\text{ bzw. }}\,\,t{\sqrt {\vert {D}\vert }}={\frac {1}{4}}{\text{ und }}t={\frac {1}{4{\sqrt {\vert {D}\vert }}}}.}$

Damit ist die zweite Koordinate gleich ${\displaystyle {}{\frac {\sqrt {\vert {D}\vert }}{4}}-{\frac {1}{4{\sqrt {\vert {D}\vert }}}}}$. Der Umkreismittelpunkt ist also

${\displaystyle \left({\frac {1}{2}},\,{\frac {\sqrt {\vert {D}\vert }}{4}}-{\frac {1}{4{\sqrt {\vert {D}\vert }}}}\right)}$

${\displaystyle {}{\frac {1}{4}}+{\left({\frac {\sqrt {\vert {D}\vert }}{4}}-{\frac {1}{4{\sqrt {\vert {D}\vert }}}}\right)}^{2}={\frac {1}{4}}+{\frac {\vert {D}\vert }{16}}+{\frac {1}{16\vert {D}\vert }}-{\frac {1}{8}}={\frac {1}{16}}{\left(2+\vert {D}\vert +{\frac {1}{\vert {D}\vert }}\right)}\,.}$