Question

A circle has a center that falls on the line `y = 6/7x +7` and passes through `( 7 ,8 )` and `(3 ,1 )`. What is the equation of the circle?

Answer 1

Equation of the circle:

`(x-1/4)^2+(y-101/14)^2=36205/784`

Explanation:

`k=6/7h+7" "`first equation
`(x-h)^2+(y-k)^2=r^2`

`(7-h)^2+(8-k)^2=r^2" "`second equation
`(3-h)^2+(1-k)^2=r^2" "`third equation

We have 3 equations with 3 unknows h, k, r

Using second and third equations, we can eliminate r

`r^2=r^2`
`(7-h)^2+(8-k)^2=(3-h)^2+(1-k)^2`

`49-14h+h^2+64-16k+k^2=9-6h+h^2+1-2k+k^2`
`103-8h-14k=0`

Our fourth equation
`8h+14k=103`

use the first equation now with the fourth equation
`8h+14k=103`
`8h+14(6/7h+7)=103`
`8h+12h+98=103`
`20h=5`
`h=1/4`

solve k:

`k=6/7h+7`
`k=(6/7)(1/4)+7`
`k=3/14+7`
`k=(3+98)/14`
`k=101/14`

solve for r:

`(3-h)^2+(1-k)^2=r^2`
`(3-1/4)^2+(1-101/14)^2=r^2`
`(11/4)^2+(-87/14)^2=r^2`
`121/16+7569/196=r^2`
`r^2=36205/784`

Equation of the circle:

`(x-1/4)^2+(y-101/14)^2=36205/784`

kindly see the graph of line `y=6/7x+7` and the circle `(x-1/4)^2+(y-101/14)^2=36205/784`

graph{((x-1/4)^2+(y-101/14)^2-36205/784)(y-6/7x-7)=0[-30,30,-15,15]}

God bless....I hope the explanation is useful.