Question

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

Answer 1

The equation of the circle is `(x-11/18)^2+(y-35/6)^2=2405/162`

Explanation:

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

`C=((4+1)/2,(4+2)/2)=(5/2,3)`

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

The slope of the line perpendicular to `AB` is `=-3/2`

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

`y-3=-3/2(x-5/2)`

`y=-3/2x+15/4+3=-3/2x+27/4`

The intersection of this line with the line `y=3x+4` gives the center of the circle.

`3x+4=-3/2x+27/4`

`3x+3/2x=27/4-4`

`9/2x=11/4`

`x=11/18`

`y=3*(11/18)+4=35/6`

The center of the circle is `(11/18,35/6)`

The radius of the circle is

`r^2=(1-11/18)^2+(2-35/6)^2`

`=(7/18)^2+(-23/6)^2`

`=4810/324=2405/162`

The equation of the circle is

`(x-11/18)^2+(y-35/6)^2=2405/162`
graph{((x-11/18)^2+(y-35/6)^2-2405/162)(y-3x-4)(y+3/2x-27/4)=0 [-9.25, 10.75, -2.56, 7.44]}