Question

A circle has a center that falls on the line `y = 8/7x +2` and passes through `( 2 ,1 )` and `(3 ,6 )`. What is the equation of the circle?

Answer 1

`C(70/47,174/47)` radius`=root2(16705/2209)`
Equation of circumference is `(x-70/47)^2+(y-174/47)^2=16705/2209`

Explanation:

`A(2,1)` and `B(3,6)` a point `C` on the line `y=8/7x+2` has coordinates `C(x,8/7x+2)`.

The centre `C` can be found by imposing that `AC=BC`

`root2((2-x)^2+(1-8/7x-2)^2)=root2((3-x)^2+(6-8/7x-2)^2)`
By squaring boh members we get

`4-4x+x^2+1+16/7x+64/49x^2=9-6x+x^2+16-64/7x+64/49x^2`
that properly manipulated becomes
`16/7x-4x+64/7x+6x=25-5` from which
`x=70/47` and consequently
`y=8/7*70/47+2=80/47+2=174/47`.
These are the coordinates of the centre `C`
The squared radius is `R^2=(2-70/47)^2+(1-174/47)^2=16705/2209`