Observe that #A(0,8)# lies of #y#-axis #T(-6,0)# lies on #x#-axis and #O(0,0)# is origin, hence
#DeltaOAT# is a right angled triangle, whose hypotenuse is #AT# and hence centre of its circumscribed circle is mid-point of #AT# and radius is #1/2AT#.
As midpoint of #A(0,8)# and #T(-6,0)# is #((0-6)/2,(8-0)/2)# i.e. #(-3,4)# is centre of the circumscribing circle. (Note that it is possible only in a right angled triangle.)
Further as length of #AT# is #sqrt((0-(-6))^2+(8-0)^2)#
= #sqrt(6^2+8^2)=sqrt(36+64)=sqrt100=10#, hence radius of circumscribed circle is #5#.
Therefore equation of circumscribing circle is
#(x-(-3))^2+(y-4)^2=25# i.e. #(x+3)^2+(y-4)^2=25#
or #x^2+6x+9+y^2-8y+16=25#
i.e. #x^2+y^2+6x-8y=0#
graph{(x^2+y^2+6x-8y)(x^2+y^2-0.02)((x+3)^2+(y-4)^2-0.02)((x+6)^2+y^2-0.02)(x^2+(y-8)^2-0.02)=0 [-12.12, 7.88, -0.88, 9.12]}
