Question

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

Answer 1

The equation of the circle is `(x+70/27)^2+(y-122/27)^2=24401/729`

Explanation:

Let `C` be the mid point of `A=(2,1)` and `B=(3,6)`

`C=((2+3)/2,(6+1)/2)=(5/2,7/2)`

The slope of `AB` is `=(6-1)/(3-2)=(5)/(1)=5`

The slope of the line perpendicular to `AB` is `=-1/5`

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

`y-7/2=-1/5(x-5/2)`

`y=-1/5x+1/2+7/2=-1/5x+4`

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

`4/7x+6=-1/5x+4`

`4/7x+1/5x=4-6`

`27/35x=-2`

`x=-70/27`

`y=4/7*(-70/27)+6=122/27`

The center of the circle is `(-70/27,122/27)`

The radius of the circle is

`r^2=(2+70/27)^2+(1-122/27)^2`

`=(124/27)^2+(-95/27)^2`

`=24401/729`

The equation of the circle is

`(x+70/27)^2+(y-122/27)^2=24401/729`
graph{((x+70/27)^2+(y-122/27)^2-24401/729)(y-4/7x-6)(y+1/5x-4)=0 [-15.77, 12.7, -2.41, 11.83]}