Find the equation of the circle where the center is on the line 8x + 5y - 8 =0 and passing through point (2,1) and point (3,5)

Answer 1

In general: (x-a)^2 + (y-b)^2 = r^2
Center point in 8x + 5y = 8, meaning: 8a + 5b = 8. ..(1)

Circle passing (2,1):
(2-a)^2 + (1-b)^2 = r^2. ..(2)

Circle passing (3,5):
(3-a)^2 + (5-b)^2 = r^2. ..(3)

Use equations (2) and (3) to eliminate r, we now have a new equation with a and b as variables. Use this last equation along with equation (1) to find the values of a and b
And to find r, you can use either equation (2) or (3)

Answer 2

or

to avoid the messy quadratic....

the centre must lie on the right bisector of the line joining (2,1) and (3,5)

slope of that line is 4, so the slope of the right-bisector is -1/4
midpoint is (5/2 , 3)
so equation of that right bisector is
y = (-1/4)x + b
using (2,1)
1 = -1/2 + b
b = 3/2
y = (-1/4)x) + b
solving that with 8x + 5y = 8 using substitution, I got
x = -3/2 and y = 4
so the centre is (-3/2 , 4)

equation of circle:
(x + 3/2)^2 + (y-4)^2 = r^2
sub in (2,1)
r^2 = 85/4

equation:
(x + 3/2)^2 + (y-4)^2 = 85/4

check: does (3,5) satisfy this equation?
yes it does!