A circle has a center that falls on the line `y = 5/8x +6` and passes through `( 5 ,2 )` and `(3 ,4 )`. What is the equation of the circle?

Let the center of the circle be `(a,b)`

Then the equation of the circle is

`(x-a)^2+(y-b)^2=r^2`

The points on the circle are `(5,2)` and `(3.4)`

Therefore,

`(5-a)^2+(2-b)^2=r^2`

and `(3-a)^2+(4-b)^2=r^2`

So,

`(5-a)^2+(2-b)^2=(3-a)^2+(4-b)^2`

`25-10a+a^2+4-4b+b^2=9-6a+a^2+16-8b+b^2`

`4-10a-4b=-6a-8b`

`4a-4b=4`

`a-b=1` this is the first equation

`(a,b)` lies on the line `y=5/8x+6`

So, `b=5/8a+6`

So, `8b=5a+48` this is the second equation

Solving for `(a,b)` with the 2 simultaneous equations

`8b=5(b+1)+48`

`3b=53`, `=>`, `b=53/3`

`a=b+1=53/3+1=56/3`

The center of the circle is `(56/3,53/3)`

The radius is `r^2=(5-a)^2+(2-b)^2`

`r^2=(5-56/3)^2+(2-53/3)^2`

`r=sqrt((41/3)^2)+(47/3)^2)`

`=20.8`

The equation of the circle is

`(x-56/3)^2+(y-53/3)^2=20.8^2`

graph{((x-56/3)^2+(y-53/3)^2-20.8^2)(y-5/8x-6)=0 [-105.4, 105.5, -52.8, 52.7]}

Let the center of the circle be `(a,b)`

Then the equation of the circle is

`(x-a)^2+(y-b)^2=r^2`

The points on the circle are `(5,2)` and `(3.4)`

Therefore,

`(5-a)^2+(2-b)^2=r^2`

and `(3-a)^2+(4-b)^2=r^2`

So,

`(5-a)^2+(2-b)^2=(3-a)^2+(4-b)^2`

`25-10a+a^2+4-4b+b^2=9-6a+a^2+16-8b+b^2`

`4-10a-4b=-6a-8b`

`4a-4b=4`

`a-b=1` this is the first equation

`(a,b)` lies on the line `y=5/8x+6`

So, `b=5/8a+6`

So, `8b=5a+48` this is the second equation

Solving for `(a,b)` with the 2 simultaneous equations

`8b=5(b+1)+48`

`3b=53`, `=>`, `b=53/3`

`a=b+1=53/3+1=56/3`

The center of the circle is `(56/3,53/3)`

The radius is `r^2=(5-a)^2+(2-b)^2`

`r^2=(5-56/3)^2+(2-53/3)^2`

`r=sqrt((41/3)^2)+(47/3)^2)`

`=20.8`

The equation of the circle is

`(x-56/3)^2+(y-53/3)^2=20.8^2`

graph{((x-56/3)^2+(y-53/3)^2-20.8^2)(y-5/8x-6)=0 [-105.4, 105.5, -52.8, 52.7]}