Given : The center of the a circle lies on the equation of line #y=3/8x+8# and passes through points #A(a_x, a_y) = A(7,4)# and #B(b_x, b_y)=B(2,9)#
Required : Equation of the circle
Solution strategy: a) Equation of the circle centered at #O(x_c, y_c)# #=> (x+x_c)^2 + (y+y_c)^2= r^2#
The center of the circle #(x_c, y_c)=(x, 3/8x+8)#
b) From the distance formula:
#r^2 = (x-x_c)^2 + (y-y_c)^2# such that
#A=>x: x= a_x=7 and y: y=a_y=4#
#B=>x: x= b_x=2 and y: y=b_y=9#
We start with b) and writing two Distance Formula equations for A and B respectively:
#A=> r^2= (x-7)^2+(3/8x+8-4)^2#
# =(x-7)^2+(3/8x+4)^2#
#B=> r^2= (x-2)^2+(3/8x+8-9)^2#
# = (x-2)^2+(3/8x-1)^2#
Now both #A and B# lie on the circle so thy are equal to one another, i.e #r^2=r^2# thus,
# (x-7)^2+(3/8x+4)^2 = (x-2)^2+(3/8x-1)^2# Rearrange and clean up #64(x-7)^2 + (3x+32)^2 = 64(x-2)^2 + (3x-8)^2#
Expand and eliminate the square in #x# terms:
#64(x^2-14x+49) + (9x^2+192x+1024) = #
#64(x^2-4x+4) + (9x^2-48x+64)#
#cancel(73x^2)-704x+4160=cancel(73x^2)-304x+320# Solve for #x#
#x=48/5#. Now insert #x=48/5# on the equation of line #y=3/8x+8#
#y=3/8*48/5 +8 = 58/5#
Thus the center, #O(x_c, y_c)= O(48/5, 58/5)#
To verify calculate the radiuses, #bar(OA) and bar(OB)#
#(bar(OA))^2=(x_c-7)^2+(y_c-4)^2=(48/5-7)^2+(58/5-4)^2#
#(bar(OA))^2 = 169/25 + 1444/25#
#(bar(OB))^2=(x_c-2)^2+(y_c-9)^2=(48/5-2)^2+(58/5-9)^2#
#(bar(OB))^2 = 1444/25 + 169/25#
#(bar(OA))^2=(bar(OB))^2 = r^2#
Now from strategy a) we have
#(x+x_c)^2+(y+y_c)^2= r^2# replace #(x_c, y_c) and r^2# by #(48/5, 58/5) and r^2=1613/25#
#(x+48/5)^2+(y+58/5)^2= 1613/25#
