Question

A circle has a center that falls on the line `y = 2/7x +3` and passes through `( 2 ,8 )` and `(3 ,5 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x+56)^2+(y+13)^2=3805`

Explanation:

Let `C` be the mid point of `A=(2,8)` and `B=(3,5)`

`C=((2+3)/2,(8+5)/2)=(5/2,13/2)`

The slope of `AB` is `=(5-8)/(3-2)=(-3)/(1)=-3`

The slope of the line perpendicular to `AB` is `=1/3`

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

`y-13/2=1/3(x-5/2)`

`y=1/3x-5/6+13/2=1/3x+17/3`

The intersection of this line with the line `y=2/7x+3` gives the center of the circle.

`1/3x+17/3=2/7x+3`

`1/3x-2/7x=3-17/3`

`1/21x=-8/3`

`x=-21*8/3=-56`

`y=2/7*(-56)+3=-13`

The center of the circle is `(-56,-13)`

The radius of the circle is

`r^2=(2+56)^2+(8+13)^2`

`=(58)^2+(21)^2`

`=3805`

The equation of the circle is

`(x+56)^2+(y+13)^2=3805`

graph{((x+56)^2+(y+13)^2-3805)(y-2/7x-3)(y-1/3x-17/3)=0 [-165.8, 101, -73.3, 60.2]}