If the points lie on a circle with radius 11 then they can be represented as
e^(i phi_1)+e^(i phi_2)+e^(i phi_3)+e^(i phi_4)=0eiϕ1+eiϕ2+eiϕ3+eiϕ4=0
or equivalently
sin phi_1+sin phi_2+sin phi_3+sin phi_4 = 0sinϕ1+sinϕ2+sinϕ3+sinϕ4=0
and
cos phi_1+cos phi_2+cos phi_3+cos phi_4 = 0cosϕ1+cosϕ2+cosϕ3+cosϕ4=0
(Here we used de Moivre's identity e^(i phi) = cos phi + i sin phieiϕ=cosϕ+isinϕ )
grouping and squaring both sides
(sin phi_1+sin phi_2)^2=(-(sin phi_3+sin phi_4))^2(sinϕ1+sinϕ2)2=(−(sinϕ3+sinϕ4))2
(cos phi_1+cos phi_2)^2=(-(cos phi_3+cos phi_4))^2(cosϕ1+cosϕ2)2=(−(cosϕ3+cosϕ4))2
and adding side by side we get at
2+2(sin phi_1 sin phi_2+cos phi_1 cos phi_2) = 2 +2(sin phi_3 sin phi_4+ cos phi_3 cos phi_4)2+2(sinϕ1sinϕ2+cosϕ1cosϕ2)=2+2(sinϕ3sinϕ4+cosϕ3cosϕ4)
or
cos(phi_2-phi_1) = cos(phi_4-phi_3)cos(ϕ2−ϕ1)=cos(ϕ4−ϕ3)
grouping now
(sin phi_2+sin phi_3)^2=(-(sin phi_1+sin phi_4))^2(sinϕ2+sinϕ3)2=(−(sinϕ1+sinϕ4))2
(cos phi_2+cos phi_3)^2=(-(cos phi_1+cos phi_4))^2(cosϕ2+cosϕ3)2=(−(cosϕ1+cosϕ4))2
we get at
cos(phi_3-phi_2) = cos(phi_4-phi_1)cos(ϕ3−ϕ2)=cos(ϕ4−ϕ1)
So we can conclude that the quadrilateral is a regular quadrilateral.
