Question 1004217
{{{drawing(300,300,-.1,.4,-.4,.1,
grid(0),
locate(0.248,-0.005,A(sqrt(6),0)),
locate(0.01,-0.245,B(0,-sqrt(6))),
line(0,-0.245,0.245,0),
line(0,0,3,-3),
locate(0.005,0.033,O(0,0)),locate(.135,-.107,M)
)}}}
Points {{{A(sqrt(6),0)}}} and {{{B(0,-sqrt(6))}}} both are at distance {{{sqrt(6)}}} from {{{O(0,0)}}} , the origin.
So, triangle {{{ABO}}} is a right isosceles triangle, with {{{45^o}}} angles at {{{a}}} and {{{B}}} .
Points {{{A}}} and {{{B}}} are on the same arc of a circle of radius {{{r}}} ,
so both are at distance {{{r}}} from the center of the circle, {{{C}}} .
Since the center is at the same distance from {{{A}}} and {{{B}}} ,
the center is on the perpendicular bisector of {{{AB}}} ,
which contains {{{M}}} , the midpoint of {{{AB}}} .
The perpendicular bisector of the base of an isosceles triangle passes through the vertex,
so the perpendicular bisector of AB is {{{OM}}} .
The distance from {{{A}}} to {{{B}}} is
{{{AB=sqrt(6+6)=sqrt(12)=2sqrt(3)}}} ,
so {{{AM=MB=2sqrt(3)/2=sqrt(3)}}} .
Adding point {{{C}}} to the sketch and drawing isosceles triangle {{{ABC}}} we get
{{{drawing(300,300,-.1,.4,-.4,.1,
grid(0),
locate(0.248,-0.005,A(sqrt(6),0)),
locate(0.01,-0.245,B(0,-sqrt(6))),
line(0,-0.245,0.245,0),
line(0,0,3,-3),
locate(0.005,0.033,O(0,0)),locate(.135,-.107,M),
line(0,-0.245,0.0518,-0.0518),
line(0.245,0,0.0518,-0.0518),
red(arc(0.0518,-0.0518,0.1,0.1,-15,105)),
locate(0.05,-0.02,C)
)}}}
{{{Angle}}}{{{ACB=120^o}}} , and as {{{ABC}}} is an isosceles triangle,
altitude {{{AC}}} splits it into two congruent {{{30-60-90}}} triangles: {{{ACM}}} and {{{BCM}}} .
{{{Angle}}}{{{BAC=30^o}}} , and {{{AC*cos(BAC)=AM}}} .
Substituting the measures we know
{{{r*cos(30^o)=sqrt(3)}}} , and since {{{cos(30^o)=sqrt(3)/2}}} ,
{{{r*(sqrt(3)/2)=sqrt(3)}}}-->{{{highlight(r=2)}}} .