Question

A circle has a center that falls on the line `y = 4/7x +6` and passes through `( 2 ,1 )` and `(3 ,5 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x+70/23)^2+(y-98/23)^2=19081/529`

Explanation:

Let `C` be the mid point of `A=(2,1)` and `B=(3,5)`

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

The slope of `AB` is `=(5-1)/(3-2)=(4)/(1)=4`

The slope of the line perpendicular to `AB` is `=-1/4`

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

`y-3=-1/4(x-2)`

`y=-1/4x+1/2+3=-1/4x+7/2`

The intersection of this line with the line `y=4/7x+6` gives the center of the circle.

`-1/4x+7/2=4/7x+6`

`1/4x+4/7x=7/2-6`

`23/28x=-5/2`

`x=-28/23*5/2=-70/23`

`y=4/7*(-70/23)+6=98/23`

The center of the circle is `(-70/23,98/23)`

The radius of the circle is

`r^2=(2+70/23)^2+(1-98/23)^2`

`=(116/23)^2+(75/23)^2`

`=19081/529`

The equation of the circle is

`(x+70/23)^2+(y-98/23)^2=19081/529`

graph{ ((x+70/23)^2+(y-98/23)^2-19081/529)(y-4/7x-6)(y+1/4x-7/2)=0 [-36.54, 36.55, -18.26, 18.28]}