Let the points are #P(8,-2) , Q( 6,2) ,R(3,-7)# and Centre of
the circle is #(a,b)# and radius is #r#
The distance from center to the given 3 points are equal.
#(8-a)^2+(-2-b)^2=(6-a)^2+(2-b)^2=(3-a)^2+(-7-b)^2#
#(8-a)^2+(-2-b)^2=(6-a)^2+(2-b)^2#
#:. 64-16a+cancela^2+4+4b+cancelb^2= 36-12a+cancela^2+4-4b+cancelb^2# or
#-4a+8b=-28 or 2b-a= -7(1)# Similarly,
#(8-a)^2+(-2-b)^2=(3-a)^2+(-7-b)^2#
#:. 64-16a+cancela^2+4+4b+cancelb^2= 9-6a+cancela^2+49+14b+cancelb^2# or
#-10a-10b= =-10 or a+b=1(2)# Adding equation(1) and (2) we
get, #3b=-6 or b =-2:. a =1-b=1-(-2)=3 :.#
Centre is #(3,-2)# Radius is #r=|PC|=sqrt ((8-a)^2+(-2-b)^2)#
#:. r= sqrt ((8-3)^2+(-2+2)^2)=5#
The equation of circle is #(x-a)^2+(y-b)^2=r^2# or
#(x-3)^2+(y+2)^2=25# [Ans]
