Question

A circle has a center that falls on the line `y = 12/7x +3` and passes through `( 9 ,5 )` and `(8 ,7 )`. What is the equation of the circle?

Answer 1

The center of the circle is `(-7/11,91/44)`
And the radius is `9.95`
The equation of the circle is `(x+7/11)^2+(y-91/44)^2=98.9`

Explanation:

Let `(a,b)` be the center of the circle and `r` the radius
The equation of a circle is `(x-a)^2+(y-b)^2=r^2`
So we apply this to the 2 points and we get

`(9-a)^2+(5-b)^2=r^2`
`(8-a)^2+(7-b)^2=r^2`
So we can combine these 2 equations
`(9-a)^2+(5-b)^2=(8-a)^2+(7-b)^2`
Developing

`81-18a+a^2+25-10b+b^2=64-16a+a^2+49-14b+b^2`

Simplifying
`81-18a+25-10b=64+49-16a-14b`
`106-113=18a-16a-14b+10b`
`-7=-2a-4b` `=>` `4b=-2a+7` this equation `1`
From the equation of the line, we get
`y=(12x)/7+3` `=>` `b=(12a)/7+3` this is equation `2`

So`-a/4+7/4=(12a)/7+3` `=>` `a=-7/11`
and `4b=-2*-7/11+7` `=>` `b=91/44`

from these we can calculate `r`
`r=sqrt((8--7/11)^2+(7-91/44)^2)=9.95`