Question

A circle has a center that falls on the line `y = 7/2x +3` and passes through `(1 ,2 )` and `(8 ,5 )`. What is the equation of the circle?

Answer 1

The equation of the circle is `(x-66/35)^2+(y-48/5)^2=71717/1225`

Explanation:

Let `C` be the mid point of `A=(1,2)` and `B=(8,5)`

`C=((1+8)/2,(5+2)/2)=(9/2,7/2)`

The slope of `AB` is `=(5-2)/(8-1)=(3)/(7)`

The slope of the line perpendicular to `AB` is `=-7/3`

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

`y-7/2=-7/3(x-9/2)`

`y=-7/3x+21/2+7/2=-7/3x+14`

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

`-7/3x+14=7/2x+3`

`7/2x+7/3x=14-3`

`35/6x=11`

`x=66/35`

`y=7/2*(66/35)+3=48/5`

The center of the circle is `(66/35,48/5)`

The radius of the circle is

`r^2=(1-66/35)^2+(2-48/5)^2`

`=(-31/35)^2+(-38/5)^2`

`=11069/1225`

The equation of the circle is

`(x-66/35)^2+(y-48/5)^2=71717/1225`

graph{((x-66/35)^2+(y-48/5)^2-71717/1225)(y-7/2x-3)(y+7/3x-14)=0 [-23.05, 12.98, -0.97, 17.06]}