Lets start by graphing the given points:
Now let's find the circumcenter:
[tex]M_{}AB=(\frac{14+4}{2},\frac{12+6}{2})[/tex][tex]M\text{AB}=(9,9)[/tex]
And the slope of AB is:
[tex]\frac{12-6}{4-14}=\frac{6}{-10}=\frac{3}{-5}=-\frac{3}{5}[/tex]
So the slope of the perpendicular line must be:
[tex]x\cdot-\frac{3}{5}=-1[/tex][tex]x=-\frac{1}{-\frac{3}{5}}[/tex][tex]x=\frac{5}{3}[/tex]
The equation of that bisector must be (from the slope-point formula):
[tex](y-9)=\frac{5}{3}(x-9)[/tex]
[tex]y-9=\frac{5}{3}x-\frac{45}{3}[/tex][tex]y-9=\frac{5}{3}x-15[/tex][tex]y=\frac{5}{3}x-15+9[/tex][tex]y=\frac{5}{3}x-6[/tex]
For the midpoint of BC we get:
[tex]MBC=(\frac{14-6}{2},\frac{2+6}{2})[/tex]
As shown in the graph.
The slope of BC is:
[tex]\frac{6-2}{14-(-6)}=\frac{4}{20}=\frac{2}{10}=\frac{1}{5}[/tex]
So the slope of the perpendicular line must be:
[tex]x\cdot\frac{1}{5}=-1[/tex][tex]x=-5[/tex]
The equation of that bisector must be:
[tex]y-4=-5(x-4)[/tex][tex]y-4=-5x+20[/tex][tex]y=-5x+24[/tex]
Now equate the bisectors to find the circumcenter:
[tex]-5x+24=\frac{5}{3}x-6[/tex][tex]-5x-\frac{5}{3}x=-6-24[/tex][tex]-\frac{20}{3}x=-30[/tex][tex]x=\frac{-30\cdot3}{-20}[/tex][tex]x=\frac{90}{20}=\frac{9}{2}[/tex]
And replacing on any of the bisectors:
[tex]y=-5x+24[/tex][tex]y=-5(\frac{9}{2})+24[/tex][tex]y=-\frac{45}{2}+24[/tex][tex]y=-\frac{45}{2}+\frac{48}{2}[/tex][tex]y=\frac{3}{2}[/tex]
So the circumcenter is:
[tex](\frac{9}{2},\frac{3}{2})[/tex]
