Let's write the center line 4y=5x+164y=5x+16 and let x=4cx=4c so the center is (4c, 5c + 4)(4c,5c+4) eliminating the messy fractions. Let's call the squared radius kk. The equation for the circle is thus
(x - 4c)^2 + (y - (5c + 4))^2 = k(x−4c)2+(y−(5c+4))2=k
Our job is to find cc and kk; our two points give us two equations:
(4 -4c)^2 + (7 - (5 c + 4))^2 = k(4−4c)2+(7−(5c+4))2=k
(2 - 4c)^2 + (5 - (5 c + 4))^2 = k(2−4c)2+(5−(5c+4))2=k
We expand and subtract, which should give us an equation for c.c.
16 - 32 c + 16c^2 + 9 - 30c + 25c^2 = k16−32c+16c2+9−30c+25c2=k
4 - 16 c + 16c^2 + 1 -10 c + 25c^2 = k4−16c+16c2+1−10c+25c2=k
20 -36 c = 020−36c=0
c = 20/36= 5/9c=2036=59
(x - 4(5/9))^2 + (y - (5(5/9) + 4))^2 = k(x−4(59))2+(y−(5(59)+4))2=k
(x-20/9)^2 + (y - 61/9)^2 = k(x−209)2+(y−619)2=k
k =(2-20/9)^2 + (5 - 61/9)^2 = 260/81k=(2−209)2+(5−619)2=26081
Our equation is
(x-20/9)^2 + (y - 61/9)^2 = 260/81(x−209)2+(y−619)2=26081
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]}
