Question

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

Answer 1

Hence the circle has a center on `y=2x+4` its coordinates are
`(a,2a+4)`

Now the distance of the two points from the center is equal hence

#sqrt[(4-a)^2+(4-2a-4)^2]=sqrt[(1-a)^2+(2-2a-4)^2]=> [(4-a)^2+(4-2a-4)^2]=[(1-a)^2+(2-2a-4)^2]=> a=11/14#

Hence the center is `(11/14,39/7)` and the radius is

`r=sqrt[(4-11/14)^2+(4-39/7)^2]=sqrt(2509)/14`

Hence the equation of the circle becomes

`(x-11/14)^2+(y-39/7)^2=(sqrt(2509)/14)^2`