Question 771666
The center of the circle is at the same distance from the 3 points.
That means it's on the perpendicular bisector of the segments connecting those 3 points.
Points (2,3) and (4,3) are on a horizontal segment with midpoint at (3,3).
The perpendicular bisector of that segment is the vertical line that passes through (3,3), with equation {{{x=3}}}.
All we need now is a second segment and its perpendicular bisector.
The midpoint of the segment connecting (6,1) and (4,3) is
({{{(6+4)/2}}},{{{(1+3)/2}}}) or (5,2).
The slope of the line connecting (6,1) and (4,3) is
{{{(3-1)/(4-6)=2/(-2)=-1}}}
The slopes of perpendicular line multiply to yield {{{-1}}}, so the slope of the perpendicular bisector to the segment connecting (6,1) and (4,3) is
{{{(-1)/(-1)=1}}}
The perpendicular bisector to the segment connecting (6,1) and (4,3) has the equation
{{{y-2=1(x-5)}}} (in point-slope dorm based on midpoint (5,2).
{{{y-2=1(x-5)}}}-->{{{y-2=x-5}}}-->{{{y=x-3}}}
The intersection of the two perpendicular bisectors found is the solution to
{{{system(x=3,y=x-3)}}}, and that is {{{system(x=3,y=0)}}}, the point (3,0),
The center of the circle is {{{highlight("(3,0)")}}}.
The radius is the distance from that point to any of the 3 given points.
For example, using (4,3),
{{{radius=sqrt((4-3)^2+(3-0)^2)}}}-->{{{radius=sqrt(1^2+3^2)}}}-->{{{radius=sqrt(1+9)}}}-->{{{highlight(radius=sqrt(10))}}}.
{{{drawing(300,300,-2,8,-5,5,
grid(1),red(circle(3,0,0.1)),
blue(circle(3,0,sqrt(10))),
blue(circle(2,3,0.1)),blue(circle(4,3,0.1)),blue(circle(6,1,0.1))
)}}}