Question

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

Answer 1

The equation of the circle is
`(x-1/2)^2+(y-99/14)^2=4225/98`

Explanation:

Let `C` be the mid point of `A=(7,8)` and `B=(3,1)`

`C=((7+3)/2,(8+1)/2)=(5,9/2)`

The slope of `AB` is `=(1-8)/(3-7)=(-7)/(-4)=7/4`

The slope of the line perpendicular to `AB` is `=-4/7`

The equation of the line passing trrough `C` and perpendicular to `AB` is

`y-9/2=-4/7(x-5)`

`y=-4/7x+20/7+9/2=-4/7x+103/14`

The intersection of this line with the line `y=1/7x+7` gives the center of the circle.

`1/7x+7=-4/7x+103/14`

`1/7x+4/7x=103/14-7`

`5/7x=5/14`

`x=5/14*7/5=1/2`

`y=1/7*1/2+7=99/14`

The center of the circle is `(1/2,99/14)`

The radius of the circle is

`r^2=(3-1/2)^2+(1-99/14)^2`

`=(5/2)^2+(-85/14)^2`

`=8450/196=4225/98`

The equation of the circle is

`(x-1/2)^2+(y-99/14)^2=4225/98`
graph{ ((x-1/2)^2+(y-99/14)^2-4225/98)(y-1/7x-7)=0 [-20.02, 20.52, -2.84, 17.44]}