Question

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

Answer 1

`36x^2+36y^2-396x-732y+1095=0`

Explanation:

As ordinates of points `(8,2)` and `(3,2)` are same, the line joining them is parallel to `x`-axis and its equation is `y=2` and the midpoint of points `(8,2)` and `(3,2)` being `(11/2,2)`, center lies on `x=11/2`.

Further center lies on the pont of intersection of `y=5/3x+1` and `x=11/2` and therefore we have

`y=5/3xx11/2+1=55/6+1=61/6` and center is `(11/2,61/6)` and as circle passes through `(3,2)` the radius is

`sqrt((3-11/2)^2+(2-61/6)^2)=sqrt(25/4+2401/36)=sqrt(2626/36)`

and hence equation of circle is

`(x-11/2)^2+(y-61/6)^2=2626/36`

or `x^2-11x+121/4+y^2-61/3y+3721/36=2626/36`

or `36x^2+36y^2-396x-732y+2184=0`

or `9x^2+9y^2-99x-183y+546=0`

graph{(9x^2+9y^2-99x-183y+546)((x-8)^2+(y-2)^2-0.1)((x-3)^2+(y-2)^2-0.1)(2x-11)((x-11/2)^2+(y-61/6)^2-0.1)(3y-5x-3)=0 [-16, 24, 0, 20]}