Let's write the center line #4y=5x+16# and let #x=4c# so the center is #(4c, 5c + 4)# eliminating the messy fractions. Let's call the squared radius #k#. The equation for the circle is thus
#(x - 4c)^2 + (y - (5c + 4))^2 = k#
Our job is to find #c# and #k#; our two points give us two equations:
#(4 -4c)^2 + (7 - (5 c + 4))^2 = k#
#(2 - 4c)^2 + (5 - (5 c + 4))^2 = k#
We expand and subtract, which should give us an equation for #c.#
# 16 - 32 c + 16c^2 + 9 - 30c + 25c^2 = k#
#4 - 16 c + 16c^2 + 1 -10 c + 25c^2 = k#
#20 -36 c = 0#
# c = 20/36= 5/9#
#(x - 4(5/9))^2 + (y - (5(5/9) + 4))^2 = k#
#(x-20/9)^2 + (y - 61/9)^2 = k#
#k =(2-20/9)^2 + (5 - 61/9)^2 = 260/81#
Our equation is
#(x-20/9)^2 + (y - 61/9)^2 = 260/81#
I'll plot after I post. Plot:
#0 = ( (x-20/9)^2 + (y - 61/9)^2 -260/81)( 5/4 x+4 -y)((x-4)^2+(y-7)^2-.1^2)( (x-2)^2+(y-5)^2-.1^2) ((x-20/9)^2 + (y - 61/9)^2 - .1^2)#
graph{0 = ( (x-20/9)^2 + (y - 61/9)^2 -260/81)( 5/4 x+4 -y)((x-4)^2+(y-7)^2-.1^2)( (x-2)^2+(y-5)^2-.1^2) ((x-20/9)^2 + (y - 61/9)^2 - .1^2) [-2.885, 8.365, 3.577, 9.2]}
