Question

A circle has a center that falls on the line `y = 3/8x +3` and passes through `( 1 ,4 )` and `(2 ,9 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-152/23)^2+(y-126/23)^2=17797/529`

Explanation:

Let `C` be the mid point of `A=(1,4)` and `B=(2,9)`

`C=((1+2)/2,(9+4)/2)=(3/2,13/2)`

The slope of `AB` is `=(9-4)/(2-1)=(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-13/2=-1/5(x-3/2)`

`y=-1/5x+3/10+13/2=-1/5x+34/5`

The intersection of this line with the line `y=3/8x+3` gives the center of the circle.

`3/8x+3=-1/5x+34/5`

`3/8x+1/5x=34/5-3`

`23/40x=19/5`

`x=19/5*40/23=152/23`

`y=3/8*152/23+3=126/23`

The center of the circle is `(152/23,126/23)`

The radius of the circle is

`r^2=(1-152/23)^2+(4-126/23)^2`

`=(129/23)^2+(-34/23)^2`

`=17797/529`

The equation of the circle is

`(x-152/23)^2+(y-126/23)^2=17797/529`
graph{((x-152/23)^2+(y-126/23)^2-17797/529)(y-3/8x-3)=0 [-8.25, 20.23, -2.34, 11.9]}