Generic eqn of a circle with centre #(a,b)# and radius #r# is is #(x-a)^2+(y-b)^2=r^2#
We are told the centre #(a,b)# lies on the line #y=7/2x+1#; so;
#b=7/2a+1=>2b=7a+2# ---- Equation [1]
We are also told that the circle passes through #(1,4)#; so
#(1-a)^2+(4-b)^2=r^2# ---- Equation [2A]
#:. 1-2a+a^2+16-8b+b^2=r^2#
#:. a^2+b^2 -2a-8b+17=r^2# ---- Equation [2]
We are also told that the circle passes through #(8,5)#; so
#(8-a)^2+(5-b)^2=r^2#
#:. 64-16a+a^2+25-10b+b^2=r^2#
#:. a^2+b^2 -16a-10b+89=r^2# ---- Equation [3]
Equation [3] - Equation [2] gives;
#-16a-2a-10b-8b+89-17=0#
#:.-18a-18b+72=0#
#:.18a+18b=72#
#:.a+b=4=>b=4-a# ---- Equation [4]
Substituting this value of b into Equation [1] gives:
#2(4-a)=7a+2#
#:. 8-2a=7a+2#
#:. 9a=6#
#:. a=2/3# Substitute into Equation [4] gives #b=4-2/3=10/3#
We know know #a=2/3# and #b=10/3#; so substitute these into Equation [2A] to find r:
#(1-2/3)^2+(4-10/3)^2=r^2#
#:.(1/3)^2+(2/3)^2=r^2#
#:.1/9+4/9=r^2#
#:.r^2=5/9#
So the required equation is #(x-2/3)^2+(y-10/3)^2=5/9#
