A circle has a center that falls on the line `y = 2/3x +7` and passes through `(5 ,7 )` and `(1 ,2 )`. What is the equation of the circle?

the centre of the circle is on the intersection point between the assigned line and the line axis of the segment that joint the two points
straight for two points:
`(y-y_a)/(y_b-y_a) = (x-x_a)/(x_b-x_a)`
`(y-7)/(2-7) = (x-5)/(1-5)`
`(y-7)/-5 = (x-5)/-4`
`(y-7)/-5 = (x-5)/-4`
`-4 (y-7) = -5 (x-5))`
`-4 y +28 = -5 x+ 25`
`-4 y = -5 x-3`
`y= 5/4 x + 3/4`

middle point M
`X_m =( X_b+X_a) /2=(5+1)/2=3`
`Y_m =( Y_b+Y_a) /2=(7+2)/2=9/2`

perpendicular line (m =-1/m') passing through M
`y-y_M= m (x-x_m)`
`y-9/2= -4/5(x-3)`
`y= -4/5 x +12/5+9/2`
`y= -4/5 x +59/10`

making the system and solving with comparison
`2/3 x +7= -4/5 x +69/10`
`22/15 x =-1/10`

`X_C=-3/22`
`Y_C=-1/11+7=76/11`

`R=sqrt((X_c-X_a)^2 + (Y_c-Y_a)^2)=sqrt((-3/22-1)^2 + (76/11-2)^2)=`...

i have no mere time....

equation of circle
`(x-x_c)^2 + (y-y_c)^2 = R^2`

Let, `C(h,k)` and `r` be the centre and radius of the reqd. circle `S.`

Since, `C` is on the line `y=2/3x+7," we have, "k=2/3h+7`.

`:. C(h,k)=C(h,2/3h+7)`.

The points `P(5,7) and Q(1,2) in S`.

`:. CP^2=r^2=CQ^2`.

`:.(h-5)^2+(2/3h+7-7)^2=(h-1)^2+(2/3h+7-2)^2`.

`:.(h^2-10h+25)+4/9h^2=h^2-2h+1+4/9h^2+20/3h+25`.

`:.-10h+2h-20/3h=1`.

`:. h=-3/44`.

`:. k=2/3h+7=2/3(-3/44)+7=153/22`.

Finally, `r^2=(h-5)^2+(2/3h)^2=(-3/44-5)^2+(-1/22)^2`.

`;. r^2=49733/1936`.

`:. S : (x-h)^2+(y-k)^2=r^2, i.e.,`

`(x+3/44)^2+(y-153/22)^2=49733/1936`.