Question

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?

Answer 1

see below

Explanation:

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`

Answer 2

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

Explanation:

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`.